Question

# Prove that ${{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}={{\left( 1+\sec A\operatorname{cosec}A \right)}^{2}}$.

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.