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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.

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