# Find the value of \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)\]

(a) \[0\]

(b) \[1\]

(c) \[2\]

(d) None of these

Answer

Verified

328.8k+ views

Hint: Use the properties of trigonometric functions and simplify the terms given in the bracket by multiplying each one of them and cancelling out the like terms with opposite signs.

We have to find the value of \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)\]. We will begin by simplifying the given expression by multiplying each of the terms in the two brackets.

Thus, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1\left( 1+\cot \theta -\csc \theta \right)+\tan \theta \left( 1+\cot \theta -\csc \theta \right)+\sec \theta \left( 1+\cot \theta -\csc \theta \right)\]

Further simplifying the equation, we get \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1+\cot \theta -\csc \theta +\tan \theta \cot \theta +\tan \theta -\tan \theta \csc \theta +\sec \theta +\sec \theta \cot \theta -\sec \theta \csc \theta \]

We know that \[\tan \theta \cot \theta =1\] and \[\sec \theta =\dfrac{1}{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta }\].

Thus, we have \[\tan \theta \csc \theta =\tan \theta \dfrac{1}{\sin \theta }=\dfrac{1}{\cos \theta }\].

Also, we have \[\sec \theta \cot \theta =\cot \theta \dfrac{1}{\cos \theta }=\dfrac{1}{\sin \theta }\].

Similarly, we get \[\sec \theta \csc \theta =\dfrac{1}{\sin \theta }\times \dfrac{1}{\cos \theta }=\dfrac{1}{\sin \theta \cos \theta }\].

Substituting all the above equations in the expansion of the given expression, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1+\cot \theta -\dfrac{1}{\sin \theta }+1+\tan \theta -\dfrac{1}{\cos \theta }+\dfrac{1}{\cos \theta }+\dfrac{1}{\sin \theta }-\dfrac{1}{\sin \theta \cos \theta }\]

Thus, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\cot \theta +\tan \theta -\dfrac{1}{\sin \theta \cos \theta }\].

We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta }\].

Thus, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\cot \theta +\tan \theta -\dfrac{1}{\sin \theta \cos \theta }=2+\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta \cos \theta }\]

Further simplifying the expression, we get \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta }-\dfrac{1}{\sin \theta \cos \theta }\].

We know the identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].

Thus, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\dfrac{1}{\sin \theta \cos \theta }-\dfrac{1}{\sin \theta \cos \theta }\].

So, we get \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2\].

Hence, the value of the expression \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)\] is \[2\], which is option (c).

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius \[1\]). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

We have to find the value of \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)\]. We will begin by simplifying the given expression by multiplying each of the terms in the two brackets.

Thus, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1\left( 1+\cot \theta -\csc \theta \right)+\tan \theta \left( 1+\cot \theta -\csc \theta \right)+\sec \theta \left( 1+\cot \theta -\csc \theta \right)\]

Further simplifying the equation, we get \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1+\cot \theta -\csc \theta +\tan \theta \cot \theta +\tan \theta -\tan \theta \csc \theta +\sec \theta +\sec \theta \cot \theta -\sec \theta \csc \theta \]

We know that \[\tan \theta \cot \theta =1\] and \[\sec \theta =\dfrac{1}{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta }\].

Thus, we have \[\tan \theta \csc \theta =\tan \theta \dfrac{1}{\sin \theta }=\dfrac{1}{\cos \theta }\].

Also, we have \[\sec \theta \cot \theta =\cot \theta \dfrac{1}{\cos \theta }=\dfrac{1}{\sin \theta }\].

Similarly, we get \[\sec \theta \csc \theta =\dfrac{1}{\sin \theta }\times \dfrac{1}{\cos \theta }=\dfrac{1}{\sin \theta \cos \theta }\].

Substituting all the above equations in the expansion of the given expression, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1+\cot \theta -\dfrac{1}{\sin \theta }+1+\tan \theta -\dfrac{1}{\cos \theta }+\dfrac{1}{\cos \theta }+\dfrac{1}{\sin \theta }-\dfrac{1}{\sin \theta \cos \theta }\]

Thus, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\cot \theta +\tan \theta -\dfrac{1}{\sin \theta \cos \theta }\].

We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta }\].

Thus, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\cot \theta +\tan \theta -\dfrac{1}{\sin \theta \cos \theta }=2+\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta \cos \theta }\]

Further simplifying the expression, we get \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta }-\dfrac{1}{\sin \theta \cos \theta }\].

We know the identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].

Thus, we have \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\dfrac{1}{\sin \theta \cos \theta }-\dfrac{1}{\sin \theta \cos \theta }\].

So, we get \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2\].

Hence, the value of the expression \[\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)\] is \[2\], which is option (c).

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius \[1\]). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

Last updated date: 31st May 2023

â€¢

Total views: 328.8k

â€¢

Views today: 2.85k

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE