Answer

Verified

449.4k+ views

Hint: Transform the whole equation in terms of \[\sin \theta \] and \[\cos \theta \] and then convert into desired form.

We have to prove that \[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}={{\left( 1+\sec A\operatorname{cosec}A \right)}^{2}}....\left( i \right)\]

Taking \[LHS\] of equation \[\left( i \right)\], we get

\[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}\]

We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]

Therefore, \[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}\]

\[={{\sin }^{2}}A+{{\sec }^{2}}A+2\sin A\sec A+{{\cos }^{2}}A+{{\operatorname{cosec}}^{2}}A+2\cos A\operatorname{cosec}A\]

Rearranging the equation, we get

\[\Rightarrow \left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A....\left( ii \right)\]

We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

Putting this value in equation \[\left( ii \right)\].

We get \[1+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A...\left( iii \right)\]

We know that \[\sec A=\dfrac{1}{\cos A}\] and \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]

We will put the values of \[\sec A\] and \[\operatorname{cosec}A\] in equation \[\left( iii \right)\].

We get, \[1+2\left( \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A} \right)\]

\[=1+2\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A} \right)+\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)\]

We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\].

Hence we get,

\[1+2\left( \dfrac{1}{\sin A\cos A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)\]

We know, \[\dfrac{1}{\sin A}=\operatorname{cosec}A\] and \[\dfrac{1}{\cos A}=\sec A\]

Hence, we get \[1+2\operatorname{cosec}A\sec A+{{\operatorname{cosec}}^{2}}A{{\sec }^{2}}A\]

We can write it as

\[{{\left( 1 \right)}^{2}}+{{\left( \operatorname{cosec}A\sec A \right)}^{2}}+2\left( \operatorname{cosec}A\sec A \right).1\]

We know that \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\]

By considering \[a=1\] and \[\operatorname{cosec}A\sec A=B\]

We finally get \[LHS={{\left( 1+\operatorname{cosec}A\sec A \right)}^{2}}\] which is equal to \[\text{RHS}\].

Hence Proved.

Note: By looking the terms of \[\operatorname{cosec}A\] and \[secA\] in \[\text{RHS}\], students convert \[\sin A\] and \[\cos A\] into \[\dfrac{1}{\operatorname{cosec}A}\] and \[\dfrac{1}{secA}\] respectively in first step only, but that creates confusion and does not give the desired results.

We have to prove that \[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}={{\left( 1+\sec A\operatorname{cosec}A \right)}^{2}}....\left( i \right)\]

Taking \[LHS\] of equation \[\left( i \right)\], we get

\[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}\]

We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]

Therefore, \[{{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}\]

\[={{\sin }^{2}}A+{{\sec }^{2}}A+2\sin A\sec A+{{\cos }^{2}}A+{{\operatorname{cosec}}^{2}}A+2\cos A\operatorname{cosec}A\]

Rearranging the equation, we get

\[\Rightarrow \left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A....\left( ii \right)\]

We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

Putting this value in equation \[\left( ii \right)\].

We get \[1+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A...\left( iii \right)\]

We know that \[\sec A=\dfrac{1}{\cos A}\] and \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]

We will put the values of \[\sec A\] and \[\operatorname{cosec}A\] in equation \[\left( iii \right)\].

We get, \[1+2\left( \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A} \right)\]

\[=1+2\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A} \right)+\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)\]

We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\].

Hence we get,

\[1+2\left( \dfrac{1}{\sin A\cos A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)\]

We know, \[\dfrac{1}{\sin A}=\operatorname{cosec}A\] and \[\dfrac{1}{\cos A}=\sec A\]

Hence, we get \[1+2\operatorname{cosec}A\sec A+{{\operatorname{cosec}}^{2}}A{{\sec }^{2}}A\]

We can write it as

\[{{\left( 1 \right)}^{2}}+{{\left( \operatorname{cosec}A\sec A \right)}^{2}}+2\left( \operatorname{cosec}A\sec A \right).1\]

We know that \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\]

By considering \[a=1\] and \[\operatorname{cosec}A\sec A=B\]

We finally get \[LHS={{\left( 1+\operatorname{cosec}A\sec A \right)}^{2}}\] which is equal to \[\text{RHS}\].

Hence Proved.

Note: By looking the terms of \[\operatorname{cosec}A\] and \[secA\] in \[\text{RHS}\], students convert \[\sin A\] and \[\cos A\] into \[\dfrac{1}{\operatorname{cosec}A}\] and \[\dfrac{1}{secA}\] respectively in first step only, but that creates confusion and does not give the desired results.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

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

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE