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Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

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Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry: Free PDF Download

CBSE Class 10 Maths Chapter 8 Important Questions revolve around the concept of trigonometric equations at its base. The Class 10 Maths Ch 8 Important Questions by Vedantu come with all the solutions that are drafted in a way, to provide you with a maximum understanding of the subject and figure out the different aspects of trigonometry. The important questions include questions from all the key concepts like basic trigonometry, opposite & adjacent sides in a right-angled triangle, basic trigonometric ratios, and standard values of trigonometric ratios and complementary trigonometric ratios. 


The solutions to these important questions are drafted in an easy-to-understand method. Additionally, the solutions contain step-wise explanations. Download the Class 10 important questions PDF to ensure a deeper understanding of the topic.

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Study Important Questions for Class 10 Mathematics Chapter 8 - Introduction to Trigonometry

1. If xcosθysinθ=a,xsinθ+ycosθ=b, Prove that x2+y2=a2+b2.

Ans: 

Given:

xcosθysinθ=a        …… (1)

xsinθ+ycosθ=b        …… (2)

Squaring and adding the equation (1) and (2) on both sides.

x2cos2θ+y2sin2θ2xycosθsinθ+x2sin2θ+y2cos2θ+2xycosθsinθ=a2+b2

x2(cos2θ+sin2θ)+y2(sin2θ+cos2θ)=a2+b2

x2+y2=a2+b2

x2+y2=a2+b2

Hence proved.


2. Prove that sec2θ+cosec2θ Can Never Be Less Than 2.

Ans: 

Given: sec2θ+cosec2θ 

We know that, sec2θ=1+tan2θ and cosec2θ=1+cot2θ.

sec2θ+cosec2θ=1+tan2θ+1+cot2θ

sec2θ+cosec2θ=2+tan2θ+cot2θ

Therefore, sec2θ+cosec2θ can never be less than 2.

Hence proved.


3. If sinφ=12, show that 3cosφ4cos3φ=0.

Ans: 

Given: sinφ=12

We know that sin30=12.

While comparing the angles of sin, we get 

φ=30

Substitute φ=30 to get 

3cosφ4cos3φ=3cos(30)4cos3(30)

3(32)4(338)0

Therefore, 3cosφ4cos3φ=0.

Hence proved.


4. If 7sin2φ+3cos2φ=4, then Show that tanφ=13.

Ans: 

Given: 7sin2φ+3cos2φ=4

We know that, sin2φ+cos2φ=1 and tanθ=sinθcosθ

Then, 7sin2φ+3cos2φ=4(sin2φ+cos2φ)

7sin2φ4sin2φ=4cos2φ3cos2φ

3sin2φ=cos2φ

sin2φcos2φ=13

tan2φ=13

tanφ=13

tanφ=13

Hence proved.


5. If cosφ+sinφ=2cosφ, Prove that cosφsinφ=2sinφ.

Ans: 

Given: cosφ+sinφ=2cosφ

Squaring on both sides, we get

(cosφ+sinφ)2=2cos2φ

cos2φ+sin2φ+2cosφsinφ=2cos2φ

sin2φ=2cos2φcos2φ2cosφsinφ

sin2φ=cos2φ2cosφsinφ

Add sin2φ on both sides

2sin2φ=cos2φ2cosφsinφ+sin2φ

2sin2φ=(cosφsinφ)2 

cosφsinφ=2sinφ

Hence proved.


6. If tanA+sinA=m and tanAsinA=n, then Show that m2n2=4mn.

Ans: 

Given: 

tanA+sinA=m           …… (1)

tanAsinA=n            …… (2)

Now to prove m2n2=4mn.

Take left-hand side

m2n2=(tanA+sinA)2(tanAsinA)2

tan2A+sin2A+2tanAsinAtan2Asin2A+2tanAsinA

4tanAsinA         

m2n2=4tanAsinA      …… (3)

Now take right-hand side

 4mn=4(tanA+sinA)(tanAsinA)

4tan2Asin2A4sin2Acos2Asin2A

4sin2A(1cos2A1)4sinAsec2A1

4sinAtan2A4sinAtanA

Hence, 4mn=4tanAsinA

m2n2=4mn

Hence proved. 


7. If secA=x+14x, then prove that secA+tanA=2x or 12x.

Ans: 

Given: secA=x+14x

Squaring on both sides.

sec2A=(x+14x)2

We know that, sec2A=1+tan2A

1+tan2A=(x+14x)2

tan2A=(x+14x)21

tan2A=x2+116x2+121x2+116x212(x14x)2

Taking square root on both sides,

tanA=±(x14x).

Now, find secA+tanA

If tanA=x14x means

secA+tanA=x+14x+x14x2x

secA+tanA=2x

And if tanA=x+14x means

secA+tanA=x+14xx+14x24x12x

secA+tanA=12x

Hence proved.


8. If A,Bare Acute Angles and sinA=cosB, then Find the Value of A+B.

Ans: 

Given: sinA=cosB

We know that sinA=cos(90A)

While comparing the values to get

cosB=cos(90A)

B=90AA+B=90

A+B=90.


9. Evaluate the Following Questions:

a. Solve for ϕ, if tan5ϕ=1.

Ans: 

Given: tan5ϕ=1

We know that, tan1(1)=45

5ϕ=tan1(1)45

5ϕ=45

ϕ=4559

ϕ=9


b. Solve for φ, if sinφ1+cosφ+1+cosφsinφ=4.

Ans: 

Given: sinφ1+cosφ+1+cosφsinφ=4

sin2φ+1+cos2φ+2cosφsinφ(1+cosφ)=4

sin2φ+(1+cosφ)2sinφ(1+cosφ)=4

2+2cosφsinφ(1+cosφ)=4

2(1+cosφ)sinφ(1+cosφ)=4

2sinφ=4

sinφ=12

We know that, sin30=12

sinφ=sin30φ=30

φ=30.


10. If cosαcosβ=m and cosαsinβ=n, show that (m2+n2)cos2β=n2.

Ans: 

Given: cosαcosβ=m         …… (1)

cosαsinβ=n             …… (2)

Squaring equation (1) and (2). We get,

m2=cos2αcos2β

n2=cos2αsin2β

Now to prove (m2+n2)cos2β=n2,

Take left-hand side, 

(m2+n2)cos2β=(cos2αcos2β+cos2αsin2β)cos2β

=(cos2αsin2β+cos2αcos2βcos2βsin2β)cos2β

=cos2α(1cos2βsin2β)cos2β

=cos2αsin2β

=n2

(m2+n2)cos2β=n2

Hence proved.


11. If 7cosecφ3cotφ=7, then prove that 7cotφ3cosecφ=3.

Ans: 

Given: 7cosecφ3cotφ=7

Then prove that, 7cotφ3cosecφ=3

7cosecφ3cotφ=7

Squaring on both sides, we get

49cosec2φ+9cot2φ42cosecφcotφ=49

We know that, cosec2φ=1+cot2φ and cot2φ=cosec2φ1.

49(cot2φ+1)+9(cosec2φ1)42cosecφcotφ=49

49cot2φ+49+9cosec2φ92(3cosecφ7cotφ)=49

(7cotφ3cosecφ)2=4949+9

(7cotφ3cosecφ)2=9

Take square root on both sides, we get

7cotφ3cosecφ=3

Hence proved.


12. Prove that 2(sin6φ+cos6φ) 3(sin4φ+cos4φ)+1=0.

Ans: 

Given: 2(sin6φ+cos6φ) 3(sin4φ+cos4φ)+1=0

Let us take left-hand side,

2(sin6φ+cos6φ) 3(sin4φ+cos4φ)+1=2((sin2φ)3+(cos2φ)3)3((sin2φ)2+(cos2φ)2)+1

=2[(sin2φ+cos2φ)33sin2φcos2φ(sin2φ+cos2φ)]3[(sin2φ+cos2φ)22sin2φcos2φ]+1

=2[13sin2φcos2φ]3[12sin2φcos2φ]+1

=26sin2φcos2φ3+6sin2φcos2φ+1

=1+1

=0

Therefore, 2(sin6φ+cos6φ) 3(sin4φ+cos4φ)+1=0.

Hence proved.


13. If tanθ=56 and θ=ϕ=90. What is the value of cotϕ.

Ans: 

Given: tanθ=56 and θ=ϕ=90

We know that, tanθ=1cotθ.

cotϕ=1tanϕ

=15/6

=65 

cotϕ=65.


14. What is the Value of tanφ in terms of sinφ ?

Ans: 

Given: tanφ

We know that, tanφ=sinφcosφ and cos2φ+sin2φ=1

tanφ=sinφcosφ

tanφ=sinφ1sin2φ


15. If secφ+tanφ=4, Find the Value of sinφ, cosφ.

Ans: 

Given: secφ+tanφ=4

1cosφ+sinφcosφ=4

1+sinφcosφ=4

1+sinφ=4cosφ

Squaring on both sides.

(1+sinφ)2=(4cosφ)2

1+2sinφ+sin2φ=16cos2φ

1+2sinφ+sin2φ=16(1sin2φ)

1+2sinφ+sin2φ=1616sin2φ

17sin2φ+2sinφ15=0

17sin2φ+17sinφ15sinφ15=0

17sinφ(sinφ+1)15(sinφ+1)=0

(sinφ+1)(17sinφ15)=0

If sinφ+1=0

Hence, sinφ=1 is not possible.

Then, 17sinφ15=0

sinφ=1517

Now find cosφ

1+sinφcosφ=4

Substitute the value of sinφ

1+1517=4cosφ

3217=4cosφ

cosφ=3217(4)817

cosφ=817


Short Answer Questions (2 Marks)

1. In ΔABC , Right Angled at B,AB=24cm,BC=7cm.

Determine the Following Equations:    

(i) sinA,cosA

Ans: 

Let us draw a right-angled triangle ABC, right angled at B.


A right triangle ABC with AB=24cm and BC=7cm


Using Pythagoras theorem, find AC.

AC2=AB2+BC2

=(24)2+(7)2

=576+49

=625

AC=25cm

Then,

sinA=BCAC=725

sinA=725

 cosA=ABAC=2425

cosA=2425


(ii) sinC,cosC

Ans: 

Let us draw a right-angled triangle ABC, right angled at B.


A right triangle ABC


Using Pythagoras theorem, find AC.

AC2=AB2+BC2

=(24)2+(7)2

=576+49

=625

AC=25cm

Then,

sinC=ABAC=2425       

sinC=2425

cosC=BCAC=725

cosC=725


2. In Adjoining Figure, Find the Value of tanPcotR.     


A right triangle ABC with AB=24cm and BC=7cm - (2)


Ans: 

Using Pythagoras theorem,

PR2=PQ2+QR2

(13)2=(12)2+QR2

QR2=16914425

QR=5cm

Then find tanPcotR ,

First find the value of tanP

tanP=sideopp.toPsideadj.toP=QRPQ=512

tanP=512

Now find the value of cotR

We know that, tanR=1cotR

For that we need to first find the value of tanR

tanR=sideopp.toRsideadj.toR=PQQR=125

cotR=512

Then, 

tanPcotR=QRPQQRPQ

5125120

tanPcotR=0.


3. If sinA=34, Calculate the Value of cosA and tanA.                                                         


A right triangle ABC with AC=4k and BC=3k


Ans: 

Given that the triangle ABC in which B=90

Let us take BC=3k and AC=4k

Then using Pythagoras theorem,

AB=(AC)2(BC)2

 (4k)2(3k)2

 16k9k

k7

AB=k7

Calculate the value of cosA 

cosA=ABAC=k74k=74

cosA=74

And calculate the value of tanA

tanA=BCAB=3kk7=37

tanA=37


4. Given 15cotA=8, Find the Values of sinA and secA.                                                           

Ans: 

Given: 15cotA=8

Let us assume a triangle ABC in which B=90

Then,

15cotA=8

cotA=815

Since cotA=adjhyp=ABBC.

Let us draw the triangle.


A right triangle ABC with AB=8k and BC=15k


Now, AB=8k and BC=15k.

Using Pythagoras theorem, find the value of AC.

AC=(AB)2+(BC)2

(8k)2+(15k)2

64k2+225k2

289k2

17k

AC=17k

Now, find the values of sinA and secA.

sinA=BCAC=15k17k=1517

sinA=1517

secA=ACAB=17k8k=178

 secA=178


5. If A and B are Acute Angles Such That cosA=cosB, then show that A=B


A right triangle ABC right angled at C


Ans: 

Given: cosA=cosB

In right triangle ABC,

cosA=sideadj.Ahyp.=ACAB  …… (1)

And, cosB=sideadj.Bhyp.=BCAB …… (2)

Then, cosA=cosB

Now, equate equation (1) and (2).

ACAB=BCAB

AC=BC

A=B.

Therefore, Angles opposite to equal sides are equal.

Hence proved.


6. State Whether the Following are True or False. Justify Your Answer.           

  1.  The Value of tanA is Always Less than 1.

Ans: 

False because sides of a right triangle may have any length, so tanA may have any value. For example, tanA=BCAB=1510=32=1.5


  1.  secA=125 for Some Value of Angle A.

Ans:

True as secA is always greater than 1. For example, secA=hyp.sideadj.A . As hypotenuse will be the largest side. So, it is true.


  1.  cosA is the Abbreviation Used for the Cosecant of Angle A.

Ans:

False as cosA is the abbreviation of cosineA. Because cosA means cosine of angle A and cosecA means cosecant of angle A.


  1.  cotA is the Product of cotand A.

Ans:

False as cotA is not the product of cot and A. cot without A doesn’t have meaning. 


  1.  sinθ=43 for Some Angle θ.

Ans:  

False as sinθ cannot be greater than 1. For example, sinθ=sideopp.θhyp. Since the hypotenuse is the largest side. So, sinθ will be less than 1.


7. Evaluate the Following Equations: 

i. sin18cos72

Ans:

Given: sin18cos72

We know that, sin(90θ)=cosθ

Then, 

sin18cos72=sin(9072)cos72=cos72cos72=1                   

sin18cos72=1


ii. tan26cot64

Ans:

Given: tan26cot64

We know that, tan(90θ)=cotθ

Then,

tan26cot64=tan(9064)cot64=cot64cot64=1

tan26cot64=1


iii. cos48sin42

Ans:

Given: cos48sin42

We know that, cos(90θ)=sinθ

Then, 

cos(9042)sin42

sin42sin420

 cos48sin42=0


iv. cosec31sec59

Ans:  

Given: cosec31sec59

We know that, cosec(90θ)=secθ

Then,

cosec((9059)sec59

sec59sec590

cosec31sec59=0


1. Show that the Following Equations:                       

  1. tan48tan23tan42tan67=1

Ans:

Given: tan48tan23tan42tan67=1

We know that, tan(90θ)=cotθ.

Now let us take left-hand side,

tan48tan23tan42tan67

tan(9042)tan(9067)tan42tan67

cot42cot67tan42tan67

1tan42.1tan67.tan42tan67

1 is equal to R.H.S

tan48tan23tan42tan67=1

Hence proved.


(ii) cos38cos52sin38sin52=0

Ans: 

Given: cos38cos52sin38sin52=0

We know that, cos(90θ)=sinθ

Now let us take left-hand side, 

cos38cos52sin38sin52

cos(9052)cos(9038)sin38sin52

sin52sin38sin38sin52

0 is equal to R.H.S

cos38cos52sin38sin52=0

Hence proved.


2. If  tan2A=cot(A18) where 2A is an Acute Angle, Find the Value of A

Ans: 

Given: tan2A=cot(A18)

We know that, cot(90θ)=tanθ

Then,

cot(902A)=cot(A18)

Now equalise the angles,

902A=A18

2AA=1890

3A=108

A=1083

A=36


3. If tanA=cotB, then Prove That A+B=90.                                                            

Ans: 

Given: tanA=cotB

We know that, cot(90θ)=tanθ

Then,

cot(90A)=cotB

Now equalise the angles,

90A=B

A+B=90

A+B=90

Hence proved.


4. If sec4A=cosec(A20), Where 4A is an Acute Angle, Then Find the Value of A.

Ans:  

Given: sec4A=cosec(A20)

We know that, cosec(90θ)=secθ

Then,

cosec(904A)=cosec(A20)

Now equalise the angles,

904A=A20

4AA=2090

5A=110

A=1105

A=22

Hence proved.


5. If A,B and C are Interior Angles of a ΔABC, then Show That sin(B+C2)=cosA2.

Ans: 

Given: A,B and C are interior angles of a ΔABC.

We know that, A+B+C=180.

Let us consider, 

A+B+C2=90

B+C2=90A2

Multiply sin on both sides,

sin(B+C2)=sin(90A2)

sin(B+C2)=cosA2

Hence proved.


6. Express sin67+cos75 in terms of trigonometric ratios of angles between  0 and 45.                                                   

Ans: 

Given: sin67+cos75.

We know that, sin(90θ)=cosθ and cos(90θ)=sinθ.

Then, 

sin67+cos75=sin(9023)+cos(9015)

=cos23+sin15

cos23+sin15 is the required value.


7. Express the Trigonometric Ratios sinA,secA and tanA in Terms of cotA

Ans: 

Find the value of sinA in terms of cotA.

By using identity cosec2Acot2A=1.

Then, use cosecA=1sinA

cosec2A=1+cot2A

1sin2A=1+cot2A

sin2A=11+cot2A

sinA=11+cot2A

sinA=11+cot2A

Now find the value for secA in terms of cotA.

Using identity sec2Atan2A=1

sec2A=1+tan2A

sec2A=1+1cot2A

sec2A=cot2A+1cot2A

secA=1+cot2AcotA 

secA=1+cot2AcotA

And find the value for tanA in terms of cotA

By trigonometric ratio property, tanA=1cotA

Hence, tanA=1cotA

Therefore, sinA,secA and tanA are founded in terms of cotA. 


8. Write the Other Trigonometric Ratios of A in Terms of secA.                        

Ans: 

Find the value of sinA in terms of secA

By using identity,sin2A+cos2A=1

sin2A=1cos2A

sin2A=11sec2A

sin2A=sec2A1sec2A

sinA=sec2A1secA

sinA=sec2A1secA

Now find the value for cosA in terms of secA,

By trigonometric ratio property, cosA=1secA

cosA=1secA

Find the value for tanA in terms of secA,

By using identity, sec2Atan2A=1

tan2A=sec2A1

tanA=sec2A1

tanA=sec2A1

Find the value for cosecA in terms of secA

By trigonometric ratio property, cosecA=1sinA

cosecA=1sinA

Substitute the value of sinA=sec2A1secA.

cosecA=1sec2A1secA

cosecA=secAsec2A1

cosecA=secAsec2A1

Finally, find the value for cotA in terms of secA

By trigonometric ratio property, cotA=1tanA

cotA=1tanA

Substitute the value of tanA=sec2A1

cotA=1sec2A1

cotA=1sec2A1.


9. Evaluate the Following Equations:                                    

(i) sin263+sin227cos217+cos273

Ans:

Given: sin263+sin227cos217+cos273

We know that, sin(90θ)=cosθ,cos(90θ)=sinθ and  sin2θ+cos2θ=1

Then,

sin263+sin2(9063)cos2(9073)+cos273

sin263+cos263sin273+cos273         

111                      

sin263+sin227cos217+cos273=1         


(ii) sin25cos65+cos25sin65

Ans:  

Given: sin25cos65+cos25sin65

We know that, sin(A+B)=sinAcosB+cosAsinB

Then,

sin(25+65)

sin90                    

1

sin25cos65+cos25sin65=1


10. Show that Any Positive Odd Integer Is of the Form 6q + 1, or 6q + 3,or 6q + 5, where q is some integer.                                                      

Ans: 

Let a be any positive integer and b= 6

Then, by Euclid’s algorithm,

a=6q+r for some integer q0, and r=0,1,2,3,4,5 because 0r< 6.

Therefore, a=6q or 6q+ 1 or 6q+ 2or 6q +3or 6q+ 4or 6q+ 5

Also, 6q+1=2×3q+1=2k1+1, where k1 is a positive integer

6q+3=(6q+2)+1=2(3q+1)+1=2k2+ 1,Where k2is an integer

6q+5=(6q+4)+1=2(3q+2)+1=2k3+1, where k3 is an integer

Clearly, 6q+1,6q+3,6q+5 are of the form 2k+ 1,where kan integer is.

Therefore, 6q+1,6q+3,6q+5 are not exactly divisible by 2.

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q+1, or 6q+3, or6q+5.


11. An Army Contingent of 616 Members are to March Behind an Army Band of 32 Members in a Parade. The Two Groups Are to March in the Same Number of Columns. What Is the Maximum Number of Columns in Which They Can March?

Ans: 

We have to find the HCF(616, 32) to find the maximum number of columns in which they can march. 

To find the HCF, we can use Euclid’s algorithm.

616=32×19+8

32=8×4+0

Hence,  HCF(616, 32) is 8.

Therefore, they can march in 8 columns each.


12. Use Euclid’s Division Lemma to Show That the Square of Any Positive Integer Is Either of Form 3mor 3m+1 for some integer m

[Hint: Let x be any positive integer then it is of the form3q,3q+1 or 3q+2 . Now square each of these and show that they can be rewritten in the form 3mor 3m+1.]

Ans: 

Let a be any positive integer and b=3.

Then a=3q+r for some integer q0

And r=0,1,2 because 0r<3

Therefore, a=3q or 3q+1 or 3q+2

Or,

a2=(3q)2 or (3q+1)2 or (3q+2)2

a2=(9q)2 or 9q2+6q+1 or 9q2+12q+4

a2=3×(3q)2 or 3(3q2+2q)+1 or 3(3q2+4q+1)+1

a=3k1 or 3k2+1 or 3k3+1

Where k1,k2,k3 are some positive integers 

Hence, it can be said that the square of any positive integer is either of the form 3mor 3m+1.


Short Answer Questions (3 Marks)

1. Given secθ=1312, Calculate the Values for All Other Trigonometric Ratios. 


A triangle ABC right angled at B and angle A is 𝛳


Ans: 

Given:  secθ=1312

Let us consider a triangle ABC in which A=θ and B=90

Let AB=12k and AC=13k

Then, find the value of BC

BC=(AC)2(AB)2

(13k)2(12k)269k2144k225k25k

BC=5k

Since, secθ=1312

Similarly, 

sinθ=BCAC=5k13k=513

cosθ=ABAC=12k13k=1213

tanθ=BCAB=5k12k=512

cotθ=ABBC=12k5k=125

cosecθ=ACBC=13k5k=135


2. If cotθ=78, then Evaluate the Followings Equations:                                                                                        

i. (1+sinθ)(1sinθ)(1+cosθ)(1cosθ)

Ans:

Given: cotθ=78

Let us consider a triangle ABC in which A=θ and B=90

Then, AB=7k and BC=8k

Using Pythagoras theorem, find AC

AC=(BC)2+(AB)2

(8k)2+(7k)2

64k2+49k2

113k2

113k

AC=113k


A triangle ABC right angled at B and angle A is 𝛳, sides AB=7k, BC=8k and AC=113k


Now find the value of trigonometric ratios.

sinθ=BCAC=8k113k=8113 and cosθ=ABAC=7k113k=7113.

Then, 

(1+sinθ)(1sinθ)(1+cosθ)(1cosθ)

We know that, cos2θ+sin2θ=1

Then,

1sin2θ1cos2θcos2θsin2θ

491136411349113113644964

(1+sinθ)(1sinθ)(1+cosθ)(1cosθ)=4964


ii. cot2θ

Ans: 

Given: cot2θ

We know that, cotθ=cosθsinθ

Then,

cos2θsin2θ49113641134964

cot2θ=4964

Hence, (1+sinθ)(1sinθ)(1+cosθ)(1cosθ) and cot2θ are same.


3. If 3cotA=4, then show that  1tan2A1+tan2A=cos2Asin2A.              


A triangle ABC right angled at B and sides AB=4k, BC=3k and AC=5k


Ans: 

Given: 3cotA=4

Let us consider a triangle ABC in which  B=90

Then, 3cotA=4 

cotA=43

Let AB=4k and BC=3k

Using Pythagoras theorem, find AC

AC=(BC)2+(AB)2

 (3k)2+(4k)2=16k2+9k2

25k2=5k

AC=5k

Now find the value of trigonometric ratios.

sinA=BCAC=3k5k=35  , cosA=ABAC=4k5k=45 and tanA=BCAB=3k4k=34.

To prove: 1tan2A1+tan2A=cos2Asin2A

Let us take left-hand side 

L.H.S =1tan2A1+tan2A

Substitute the value of tanA.

19161+916=1625925

725

1tan2A1+tan2A=725

Then,

R.H.S =cos2Asin2A

(45)2(35)2=1625925

725

cos2Asin2A=725.

It shows that L.H.S=R.H.S

1tan2A1+tan2A=cos2Asin2A

Hence proved.


4. In ΔABC Right Angles at B, if A=13, then Find Value of the Following Equations:                                  

(i) sinAcosC+cosAsinC


A triangle ABC right angled at B and sides AB=3k, BC=k and AC=2k


Ans: 

Let us consider a triangle ABC in which  B=90

Let BC=kand AB=3k

Then, using Pythagoras theorem find AC

AC=(BC)2+(AB)2

(k)2+(3k)2k2+3k24k22k

AC=2k

Now find the value of trigonometric ratios.

sinA=BCAC=k2k=12 and cosA=ABAC=3k2k=32

For C, adjacent =BC, opposite =AB, and hypotenuse =AC

sinC=ABAC=3k2k=32 and cosA=BCAC=k2k=12

Now find the values of the following equations,

sinAcosC+cosAsinC

12×12+32×3214+34441

sinAcosC+cosAsinC=1


(ii) cosAcosCsinAsinC

Ans:

32×1212×3234340

cosAcosCsinAsinC=0


5. In ΔPQR, right angled at Q,PR+QR=25cm and PQ=5cm. Determine the Values of  sinP,cosP  and tanP.                                                                      


A triangle PQR right angled at Q and sides PQ=5cm, QR= x cm and PR=(25-x) cm


Ans: 

Given: In ΔPQR, right angled at Q

And PR+QR=25cm, PQ=5cm

Let us take QR=xcm and PR=(25x)cm

By using Pythagoras theorem, find the value of x.

RP2=RQ2+QP2

(25x)2=(x)2+(5)262550x+x2=x2+25

50x=600x=12

Hence, RQ=12cmand RP=2512=13cm

Now, find the values of sinP,cosP  and tanP.    

sinP=RQRP=1213, cosP=PQRP=513 and tanP=RQPQ=125.


6. If tan(A+B)=3 and tan(AB)=13; 0<A+B90; A>B. Find Aand B.  

Ans: 

Given: tan(A+B)=3 and tan(AB)=13.

We know that,tan60=3 and tan30=13.

Then,

tan(A+B)=tan60

A+B=60          …… (1)

tan(AB)=tan30

AB=30          …… (2)

Adding equation (1) and (2). We get,

A+B+AB=60+302A=90A=45

A=45

Put A=45 in equation (1).

A+B=60

45+B=60B=6045B=15

B=15

Hence, A=45 and B=15.


7. Choose the Correct Option. Justify Your Choice:   

(i) 9sec2A9tan2A=

  1. 1 

  2. 9 

  3. 8 

  4. 0

Ans: (B) 9

9sec2A9tan2A

9(sec2Atan2A)9×19


(ii) (1+tanθ+secθ)(1+cotθcosecθ)=

  1. 0 

  2. 1 

  3. 2 

  4. none of these

Ans: (C) 2

(1+tanθ+secθ)(1+cotθcosecθ)

(1+sinθcosθ+1cosθ)(1+cosθsinθ1sinθ)(cosθ+sinθ+1cosθ)(sinθ+cosθ1sinθ)

We know that, sin2θ+cos2θ=1

(cosθ+sinθ)2(1)2cosθ.sinθcos2θ+sin2θ+2cosθsinθ1cosθ.sinθ        

1+2cosθsinθ1cosθ.sinθ2cosθsinθcosθ.sinθ2


(iii) (secA+tanA)(1sinA)=

  1. secA 

  2. sinA 

  3. cosecA 

  4. cosA

Ans: (D) cosA

(secA+tanA)(1sinA)

(1cosA+sinAcosA)(1sinA)(1+sinAcosA)(1sinA)

We know that, 1sin2A=cos2A

1sin2AcosAcos2AcosAcosA  


(iv) 1+tan2A1+cot2A=

  1. sec2A 

  2. 1 

  3. cot2A 

  4. none of these

Ans: (D) tan2A

1+tan2A1+cot2A=sec2Atan2A+tan2Acosec2Acot2A+cot2A

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfac{rac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

Extra close brace or missing open brace.


Long Answer Questions (4 Marks)

1. Express the Trigonometric Ratios sinA,secA and tanA in Terms of cotA.  

Ans: 

Find the value for sinA in terms of cotA

By using identity cosec2Acot2A=1

Then,

cosec2A=1+cot2A

1sin2A=1+cot2A

sin2A=11+cot2A

sinA=11+cot2A

sinA=11+cot2A

Express the value of secA in terms of cotA

By using identity sec2Atan2A=1

Then,

sec2A=1+tan2A

sec2A=1+1cot2A

sec2A=1+cot2Acot2A

secA=1+cot2AcotA

secA=1+cot2AcotA

Express the value of tanA in terms of cotA

We know that, tanA=1cotA

tanA=1cotA


2. Write the Other Trigonometric Ratios of A in Terms of secA.                 

Ans: 

Express the value of sinA in terms of secA

By using identity, sin2A+cos2A=1

sin2A=1cos2A11sec2Asec2A1sec2A

sinA=sec2A1secA

sinA=sec2A1secA

Express the value of cosA in terms of secA

We know that, cosA=1secA

cosA=1secA

Express the value of tanA in terms of secA

By using identity sec2Atan2A=1

Then,

tan2A=sec2A1

tanA=sec2A1

tanA=sec2A1

Express the value of cosecA in terms of secA

We know that, cosecA=1sinA

Then,

Substitute the value of sinA

cosecA=1sec2A1secAsecAsec2A1

cosecA=secAsec2A1

Express the value of cotA in terms of secA

We know that, cotA=1tanA

Substitute the value of tanA

cotA=1sec2A1

cotA=1sec2A1


3. Evaluate the following equations:                                                                                                       

(i). sin263+sin227cos217+cos273

Ans:

Given: sin263+sin227cos217+cos273

We know that, sin(90θ)=cosθ,cos(90θ)=sinθ and sin2θ+cos2θ=1

Then,

sin263+sin2(9063)cos2(9073)+cos273         

sin263+cos263sin273+cos273                   

111

sin263+sin227cos217+cos273=1


(ii). sin25cos65+cos25sin65

Ans: 

Given: sin25cos65+cos25sin65

We know that, sin(90θ)=cosθ,cos(90θ)=sinθ and sin2θ+cos2θ=1

sin25cos(9025)+cos25sin(9025)

sin25.sin25+cos25.cos25                   

sin225+cos225=1      

sin25cos65+cos25sin65=1


4. Choose the Correct Option. Justify Your Choice:  

(i) 9sec2A9tan2A=

  1. 1 

  2. 9 

  3. 8 

  4. 0

Ans: (B) 9

9sec2A9tan2A

9(sec2Atan2A)9×19


(ii) (1+tanθ+secθ)(1+cotθcosecθ)=

  1. 0 

  2. 1 

  3. 2 

  4. none of these

Ans: (C) 2

(1+tanθ+secθ)(1+cotθcosecθ)

(1+sinθcosθ+1cosθ)(1+cosθsinθ1sinθ)(cosθ+sinθ+1cosθ)(sinθ+cosθ1sinθ)

We know that, sin2θ+cos2θ=1

(cosθ+sinθ)2(1)2cosθ.sinθcos2θ+sin2θ+2cosθsinθ1cosθ.sinθ        

1+2cosθsinθ1cosθ.sinθ2cosθsinθcosθ.sinθ2


(iii) (secA+tanA)(1sinA)=

  1. secA 

  2. sinA 

  3. cosecA 

  4. cosA

Ans: (D) cosA

(secA+tanA)(1sinA)

(1cosA+sinAcosA)(1sinA)(1+sinAcosA)(1sinA)

We know that, 1sin2A=cos2A

1sin2AcosAcos2AcosAcosA 


(iv) 1+tan2A1+cot2A=

  1. sec2A 

  2. 1 

  3. cot2A 

  4. none of these

Ans: (D) tan2A

1+tan2A1+cot2A=sec2Atan2A+tan2Acosec2Acot2A+cot2A

sec2Acosec2A1cos2A1sin2A

sin2Acos2Atan2A.


5. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                      

(i). (cosecθcotθ)2=1cosθ1+cosθ

Ans:

Given: (cosecθcotθ)2=1cosθ1+cosθ

We know that, (ab)2=a2+b22ab, cosecθ=1sinθ and cotθ=cosθsinθ

Then, let us take left-hand side

(cosecθcotθ)2=cosec2θ+cot2θ2cosecθcotθ 

1sin2θ+cos2θsin2θ2×1sinθ.cosθsinθ1+cos2θsin2θ2cosθsin2θ

1+cos2θ2cosθsin2θ(1cosθ)2sin2θ

(1cosθ)(1cosθ)1cos2θ(1cosθ)(1cosθ)(1+cosθ)(1cosθ)

1cosθ1+cosθ=R.H.S

(cosecθcotθ)2=1cosθ1+cosθ

Hence proved.


(ii) cosA1+sinA+1+sinAcosA=2secA

Ans:

Given: cosA1+sinA+1+sinAcosA=2secA

We know that, sin2θ+cos2θ=1

Then, let us take left-hand side

cosA1+sinA+1+sinAcosAcos2A+1+sin2A+2sinA(1+sinA)cosA

cos2A+sin2A+1+2sinA(1+sinA)cosA      

1+1+2sinA(1+sinA)cosA

2+2sinA(1+sinA)cosA

2(1+sinA)(1+sinA)cosA

2cosA2secA=R.H.S

cosA1+sinA+1+sinAcosA=2secA

Hence proved.


(iii). tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

Ans:

Given: tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

We know that, a3b3=(ab)(a2+b2+ab) and sin2θ+cos2θ=1

Then, let us take L.H.S

tanθ1cotθ+cotθ1tanθ=sinθcosθ1cosθsinθ+cosθsinθ1sinθcosθ

sinθcosθ×sinθsinθcosθ+cosθsinθ×cosθcosθsinθ

sin2θcosθ(sinθcosθ)cos2θsinθ(sinθcosθ)

sin3θcos3θsinθcosθ(sinθcosθ)                     

1+sinθcosθsinθcosθ

1sinθcosθ+1

1+1sinθcosθ1+secθcosecθ = R.H.S

tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

Hence proved.


(iv) 1+secAsecA=sin2A1cosA

Ans:

Given: 1+secAsecA=sin2A1cosA

Then, let us take L.H.S 

1+secAsecA=1+1cosA1cosA

cosA+1cosA×cosA11+cosA

1+cosA×1cosA1cosA1cos2A1cosA

sin2A1cosA=R.H.S

1+secAsecA=sin2A1cosA

Hence proved.


(v) cosAsinA+1cosA+sinA1=cosecA+cotA, using the identity cosec2A=1+cot2A

Ans:

Given: cosAsinA+1cosA+sinA1=cosecA+cotA

We know that, cosec2A=1+cot2A

Then, let us take L.H.S 

cosAsinA+1cosA+sinA1

Dividing all terms by sinA

cotA1+cosecAcotA+1cosecA

cotA+cosecA1cotAcosecA+1

(cotA+cosecA)(cosec2Acot2A)(1+cotAcosecA)

(cotA+cosecA)+(cot2Acosec2A)(1+cotAcosecA)

(cotA+cosecA)(1+cotAcosecA)(1+cotAcosecA)

cotA+cosecA=R.H.S

cosAsinA+1cosA+sinA1=cosecA+cotA

Hence proved.


(vi) 1+sinA1sinA=secA+tanA

Ans:

Given: 1+sinA1sinA=secA+tanA

We know that, 1sin2θ=cos2θ and (a+b)(ab)=a2b2

Then, let us take L.H.S 

1+sinA1sinA

Let us take conjugate of the term. Then,

1+sinA1sinA×1+sinA1+sinA

(1+sinA)21sin2A

(1+sinA)2Cos2A

1+sinAcosA1cosA+sinAcosA                           

secA+tanA=R.H.S           

1+sinA1sinA=secA+tanA       

Hence proved.        


(vii) sinθ2sin3θ2cos3θcosθ=tanθ

Ans:

Given: sinθ2sin3θ2cos3θcosθ=tanθ

We know that, 1sin2θ=cos2θ

Then, let us take L.H.S

sinθ2sin3θ2cos3θcosθ=sinθ(12sin2θ)cosθ(2cos2θ1)

sinθ(12sin2θ)cosθ[2(1sin2θ)1]

sinθ(12sin2θ)cosθ(22sin2θ1)

sinθ(12Sin2θ)cosθ(12Sin2θ)

sinθcosθ=tanθ=R.H.S

sinθ2sin3θ2cos3θcosθ=tanθ

Hence proved.


(viii) (sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A

Ans:

Given: (sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A

We know that, cosec2θ=1+cot2θ and sec2θ=1+tan2θ

Then, let us take L.H.S 

(sinA+cosecA)2+(cosA+secA)2=(sinA+1sinA)2+(cosA+1cosA)2

=sin2A+1sin2A+2sinA.1sinA+cos2A+1cos2A+2cosA.1cosA

=2+2+sin2A+cos2A+1sin2A+1cos2A

=4+1+1sin2A+1cos2A

=5+cosec2A+sec2A

=5+1+cot2A+1+tan2A

=7+tan2A+cot2A=R.H.S

(sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A

Hence proved.


(ix) (cosecAsinA)(secAcosA)=1tanA+cotA

Ans:

Given: (cosecAsinA)(secAcosA)=1tanA+cotA

We know that, sin2θ+cos2θ=1

Then, let us take L.H.S 

(cosecAsinA)(secAcosA)=(1sinAsinA)(1cosAcosA)

=(1sin2AsinA)(1cos2AcosA)

=cos2AsinA×sin2AcosA

=sinA.cosA

=sinA.cosAsin2A+cos2A

Dividing all the terms by sinA.cosA

=sinA.cosAsinA.cosAsin2AsinA.cosA+cos2AsinA.cosA

=1sinAcosA+cosAsinA

=1tanA+cotA=R.H.S

(cosecAsinA)(secAcosA)=1tanA+cotA

Hence proved.


(x) (1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Ans:

Given:(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

We know that, 1+tan2θ=sec2θ and 1+cot2θ=cosec2A

Then, let us take L.H.S

(1+tan2A1+cot2A)=sec2Acosec2A

=1cos2A×sin2A1

=tan2A=R.H.S

Now, prove the Middle side 

(1tanA1cotA)2=(1tanA11tanA)2

=(1tanAtanA1tanA)2

=(1tanA(1tanA)tanA)

=(tanA)2

=tan2A=R.H.S

(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Hence proved.


6. Use Euclid’s Division Algorithm to Find the HCF of:                                     

(i) 135 and 225

Ans:  

Given: 135 and 225

We have 225>135,

So, we have to apply the division lemma to 225 and 135 to obtain

225=135×1+90

Here remainder900, again we are applying the division lemma to 135 and 90 to obtain 

135=90×1+45

Again, the remainder450, then apply the division lemma to obtain 

90=2×45+0

Since now we got remainder as zero. Here, the process get stops. 

The divisor at this stage is 45

Therefore, the HCF of 135 and 225 is 45.


(ii) 196 and 38220 

Ans:

Given: 196 and 38220

We have 38220>196

So, we have to apply the division lemma to 38220 and 196 to obtain 

38220=196×195+0

Since we get the remainder as zero, the process stops here. 

The divisor at this stage is 196

Therefore, HCF of 196 and 38220 is 196.  


(iii)  867 and 255

Ans:

Given: 867 and 255

We have 867 > 255, 

So, we have to apply the division lemma to 867and 255 to obtain 

867=255×3+102

Here remainder 1020, again apply the division lemma to255 and 102 to obtain 

255=102×2 51

Again, remainder 510, again apply the division lemma to 102 and 51 to obtain 

102=51×2+0

Since we get the remainder as zero, the process stops here. 

The divisor at this stage is 51

Therefore, HCF of 867 and 255 is 51.


7. Evaluate the Following Equations:                                                                                                              

i. sin60cos30+sin30cos60

Ans: 

Given: sin60cos30+sin30cos60

We know that, sin60=32,sin30=12 and cos60=12,cos30=32

Then,

sin60cos30+sin30cos60

=32×32+12×12

=34+14

=44=1

sin60cos30+sin30cos60=1


ii. 2tan245+cos230sin260

Ans:

Given: 2tan245+cos230sin260

We know that, tan45=1, sin60=32 and cos30=32

Then, 2tan245+cos230sin260

=2(1)2+(32)2(32)2

=2+3434=2

2tan245+cos230sin260=2


iii. cos45sec30+cosec30

Ans:

Given: cos45sec30+cosec30

We know that, cos45=12,sec30=23 and cosec30=2

=1223+2

=122+233

=12×32+23

=32×2(3+1)

=32×2(3+1)×3131

=3(31)2×2(31)

=3(31)42×22

=3268

cos45sec30+cosec30=3268


(iv) sin30+tan45cosec60sec30+cos60+cot45

Ans:

Given: sin30+tan45cosec60sec30+cos60+cot45

We know that,  sin30=12,tan45=1,cosec60=23,sec30=23,cos60=12 and cot45=1.

Then, sin30+tan45cosec60sec30+cos60+cot45

=1/2+12/32/3+1/2+1

=3+234234+3+2323

=33433+4

=33433+4×334334

=27+162432716

=4324311

sin30+tan45cosec60sec30+cos60+cot45=4324311


(v) 5cos260+4sin230tan245sin230+cos230

Ans:

Given: 5cos260+4sin230tan245sin230+cos230. Then,

=5(12)2+(33)2(1)2(12)2+(32)2

=5×14+4×43114+34

=112×6744

=6712

5cos260+4sin230tan245sin230+cos230=6712


8. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                                      

(i) (cosecθcotθ)2=1cosθ1+cosθ

Ans:

Given: (cosecθcotθ)2=1cosθ1+cosθ

We know that, (ab)2=a2+b22ab, cosecθ=1sinθ and cotθ=cosθsinθ

Then, let us take left-hand side

(cosecθcotθ)2=cosec2θ+cot2θ2cosecθcotθ 

1sin2θ+cos2θsin2θ2×1sinθ.cosθsinθ1+cos2θsin2θ2cosθsin2θ

1+cos2θ2cosθsin2θ(1cosθ)2sin2θ

(1cosθ)(1cosθ)1cos2θ(1cosθ)(1cosθ)(1+cosθ)(1cosθ)

1cosθ1+cosθ=R.H.S

(cosecθcotθ)2=1cosθ1+cosθ

Hence proved.


(ii) cosA1+sinA+1+sinAcosA=2secA

Ans:

Given: cosA1+sinA+1+sinAcosA=2secA

We know that, sin2θ+cos2θ=1

Then, let us take left-hand side

cosA1+sinA+1+sinAcosAcos2A+1+sin2A+2sinA(1+sinA)cosA

cos2A+sin2A+1+2sinA(1+sinA)cosA      

1+1+2sinA(1+sinA)cosA

2+2sinA(1+sinA)cosA

2(1+sinA)(1+sinA)cosA

2cosA2secA=R.H.S

cosA1+sinA+1+sinAcosA=2secA

Hence proved.


(iii) tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

Ans:

Given: tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

We know that, a3b3=(ab)(a2+b2+ab) and sin2θ+cos2θ=1

Then, let us take L.H.S

tanθ1cotθ+cotθ1tanθ=sinθcosθ1cosθsinθ+cosθsinθ1sinθcosθ

sinθcosθ×sinθsinθcosθ+cosθsinθ×cosθcosθsinθ

sin2θcosθ(sinθcosθ)cos2θsinθ(sinθcosθ)

sin3θcos3θsinθcosθ(sinθcosθ)                     

1+sinθcosθsinθcosθ

1sinθcosθ+1

1+1sinθcosθ1+secθcosecθ = R.H.S

tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

Hence proved.


(iv) 1+secAsecA=sin2A1cosA

Ans:

Given: 1+secAsecA=sin2A1cosA

Then, let us take L.H.S 

1+secAsecA=1+1cosA1cosA

cosA+1cosA×cosA11+cosA

1+cosA×1cosA1cosA1cos2A1cosA

sin2A1cosA=R.H.S

1+secAsecA=sin2A1cosA

Hence proved.


(v) cosAsinA+1cosA+sinA1=cosecA+cotA, using the identity cosec2A=1+cot2A

Ans:

Given: cosAsinA+1cosA+sinA1=cosecA+cotA

We know that, cosec2A=1+cot2A

Then, let us take L.H.S 

cosAsinA+1cosA+sinA1

Dividing all terms by sinA

cotA1+cosecAcotA+1cosecA

cotA+cosecA1cotAcosecA+1

(cotA+cosecA)(cosec2Acot2A)(1+cotAcosecA)

(cotA+cosecA)+(cot2Acosec2A)(1+cotAcosecA)

(cotA+cosecA)(1+cotAcosecA)(1+cotAcosecA)

cotA+cosecA=R.H.S

cosAsinA+1cosA+sinA1=cosecA+cotA

Hence proved.


(vi) 1+sinA1sinA=secA+tanA

Ans:

Given: 1+sinA1sinA=secA+tanA

We know that, 1sin2θ=cos2θ and (a+b)(ab)=a2b2

Then, let us take L.H.S 

1+sinA1sinA

Let us take conjugate of the term. Then,

1+sinA1sinA×1+sinA1+sinA

(1+sinA)21sin2A

(1+sinA)2Cos2A

1+sinAcosA1cosA+sinAcosA                           

secA+tanA=R.H.S           

1+sinA1sinA=secA+tanA       

Hence proved.        


(vii) sinθ2sin3θ2cos3θcosθ=tanθ

Ans:

Given: sinθ2sin3θ2cos3θcosθ=tanθ

We know that, 1sin2θ=cos2θ

Then, let us take L.H.S

sinθ2sin3θ2cos3θcosθ=sinθ(12sin2θ)cosθ(2cos2θ1)

sinθ(12sin2θ)cosθ[2(1sin2θ)1]

sinθ(12sin2θ)cosθ(22sin2θ1)

sinθ(12Sin2θ)cosθ(12Sin2θ)

sinθcosθ=tanθ=R.H.S

sinθ2sin3θ2cos3θcosθ=tanθ

Hence proved.


(viii) (sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A

Ans:

Given: (sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A

We know that, cosec2θ=1+cot2θ and sec2θ=1+tan2θ

Then, let us take L.H.S 

(sinA+cosecA)2+(cosA+secA)2=(sinA+1sinA)2+(cosA+1cosA)2

=sin2A+1sin2A+2sinA.1sinA+cos2A+1cos2A+2cosA.1cosA

=2+2+sin2A+cos2A+1sin2A+1cos2A

=4+1+1sin2A+1cos2A

=5+cosec2A+sec2A

=5+1+cot2A+1+tan2A

=7+tan2A+cot2A=R.H.S

(sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A

Hence proved.


(ix) (cosecAsinA)(secAcosA)=1tanA+cotA

Ans:

Given: (cosecAsinA)(secAcosA)=1tanA+cotA

We know that, sin2θ+cos2θ=1

Then, let us take L.H.S 

(cosecAsinA)(secAcosA)=(1sinAsinA)(1cosAcosA)

=(1sin2AsinA)(1cos2AcosA)

=cos2AsinA×sin2AcosA

=sinA.cosA

=sinA.cosAsin2A+cos2A

Dividing all the terms by sinA.cosA

=sinA.cosAsinA.cosAsin2AsinA.cosA+cos2AsinA.cosA

=1sinAcosA+cosAsinA

=1tanA+cotA=R.H.S

(cosecAsinA)(secAcosA)=1tanA+cotA

Hence proved.


(x) (1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Ans:

Given: (1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

We know that, 1+tan2θ=sec2θ and 1+cot2θ=cosec2A

Then, let us take L.H.S

(1+tan2A1+cot2A)=sec2Acosec2A

=1cos2A×sin2A1

=tan2A=R.H.S

Now, prove the Middle side 

(1tanA1cotA)2=(1tanA11tanA)2

=(1tanAtanA1tanA)2

=(1tanA(1tanA)tanA)

=(tanA)2

=tan2A=R.H.S

(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Hence proved.


Key Topics Covered in Chapter 8

Chapter 8 of Class 10 Mathematics consists of the discussion of the following key concepts.

  • Basic Trigonometry

  • Opposite & Adjacent Sides In A Right-Angled Triangle

  • Basic Trigonometric Ratios

  • Standard Values Of Trigonometric Ratios 

  • Complementary Trigonometric Ratios 


Significance of Important Questions of Trigonometry Class 10

Trigonometry is one of the most significant parts of Important Questions for Class 10 Maths Chapter 8. This chapter revolves around the memorization and conceptual understanding of the user’s problem-solving ability of a given space, preferably a triangle. These Class 10 Maths Trigonometry Important Questions help students to have a better understanding and application of trigonometry in the real world and clear the base around its introductory stage, to help students retain its benefits in the long run.


Did You Know?

Trigonometry has different applications and a few of them are mentioned below:

  • It is used in cartography for the creation of maps.

  • It has applications in the aviation industry and satellite systems.

  • It is also used to describe light and sound waves. 


This was the complete discussion on CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry important questions. If you are a Class 10 student we highly recommend downloading and practising the important questions on trigonometry. Practising these important questions will not only help you understand the concepts but will also help you in analysing the important topics and exam patterns. 


We wish you all the very best! Be exam ready with Vedantu!

 

Related Study Materials for Class 10 Maths Chapter 8 Introduction to Trigonometry


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FAQs on Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

1. Is Studying Chapter 8 Maths Class 10 Important Questions Helpful in Preparing for the Exams?

Studying Trigonometry Important Questions Class 10 helps you uncover the different aspects of trigonometry and teach the subject at its core. The pool of experienced teachers who have a complete idea of the marking schematics and all the necessary questions that can come in the exams are focused on providing the students with a complete kit that helps them secure maximum marks in their upcoming exams. Further, the step by step explanation of practical problems is what set our solutions apart, making it completely reliable a solution to start preparing for your exams.

2. What are the important questions in Chapter 8 Trigonometry of Class 10 Maths?

Class 10 Maths Chapter 8 is related to trigonometry and its functions. Important questions given on Vedantu for trigonometry Class 10 Chapter 8 can help students to prepare for their exams. Important questions are given according to the latest syllabus and pattern of CBSE. All questions are prepared using sample papers, previous year papers, and the latest trends in the CBSE board exams. Students can use important questions in trigonometry Class 10 to prepare for their final board exams. These solutions are available on Vedantu's official website(vedantu.com) and mobile app free of cost.

3. What is the easiest way to solve Chapter 8 Trigonometry of Class 10 Maths questions?

Students have to understand the basic concepts of Trigonometry for solving Class 10 Maths Chapter 8 questions. They have to memorize the different identities properly for solving the NCERT questions quickly. Students can also refer to the important questions in Class 10 Maths Chapter 8 available for free on Vedantu to understand the basic concepts and formulas used for solving trigonometry easily.  They can visit Vedantu to download important questions for Class 10 Maths Chapter 8 for regular practice.

4. How many exercises are there in Chapter 8 Trigonometry of Class 10 Maths?

In Trigonometry Class 10 Maths Chapter 8, there are four exercises. All exercises are based on different concepts related to trigonometry. Students need to solve all questions given in four exercises in the NCERT book to understand the concepts of trigonometry and for scoring high marks in Class 10 maths board exams. Students can practice solving trigonometry from the NCERT Solutions available on Vedantu in the offline and online mode. They can download the NCERT solutions for Class 10 Maths Chapter 8 on computers to work offline.

5. What are the different topics studied in Chapter 8 Trigonometry of Class 10 Maths?

The different topics studied in Class 10 Maths Chapter 10 include an introduction to trigonometry,  information about trigonometric ratios, trigonometric ratios of specific angles, trigonometric ratios of complementary angles, and trigonometric identities. Students will learn all topics step by step in each exercise of the chapter. They can practice NCERT Solutions given on different topics of trigonometry on Vedantu. NCERT Solutions for trigonometry Class 10 are prepared by experts to help students understand the concepts and score high marks.