Question

Prove the following identity, where the angles involved are acute angles for which the expression is defined.

${\left(\sin \left(A\right)+\cos ec\left(A\right)\right)}^2+{\left(\cos \left(A\right)+sec\left(A\right)\right)}^2=7+{\tan }^2\left(A\right)+co{t}^2\left(A\right)$

Answer 1

Solve for the required proof

Given that ${\left(\sin \left(A\right)+\cos ec\left(A\right)\right)}^2+{\left(\cos \left(A\right)+sec\left(A\right)\right)}^2=7+{\tan }^2\left(A\right)+co{t}^2\left(A\right)$

Consider the L.H.S.

${\left(\sin \left(A\right)+\cos ec\left(A\right)\right)}^2+{\left(\cos \left(A\right)+sec\left(A\right)\right)}^2$$={\sin }^2\left(A\right)+\cos e{c}^2\left(A\right)+2·\sin \left(A\right)·\cos ec\left(A\right)+{\cos }^2\left(A\right)+se{c}^2\left(A\right)+2·\cos \left(A\right)·sec\left(A\right)$ $\left[\because {\left(a+b\right)}^2=a^2+2ab+b^2\right]$

$={\sin }^2\left(A\right)+{\cos }^2\left(A\right)+\cos e{c}^2\left(A\right)+se{c}^2\left(A\right)+2·\sin \left(A\right)·\frac{1}{\sin \left(A\right)}+2·\cos \left(A\right)·\frac{1}{\cos \left(A\right)}$ ; $\left[\because \cos ec\left(A\right)=\frac{1}{\sin \left(A\right)},sec\left(A\right)=\frac{1}{\cos \left(A\right)}\right]$

$=1+1+co{t}^2\left(A\right)+1+{\tan }^2\left(A\right)+2+2$ $\left[\because {\sin }^2\left(A\right)+{\cos }^2\left(A\right)=1,1+co{t}^2\left(A\right)=\cos e{c}^2\left(A\right),1+se{c}^2\left(A\right)={\tan }^2\left(A\right)\right]$

$=7+{\tan }^2\left(A\right)+co{t}^2\left(A\right)$

$=$R.H.S

$\Rightarrow$L..H.S$=$R.H.S

Hence, it proved that ${\left(\sin A+\cos ecA\right)}^2+{\left(\cos A+secA\right)}^2=7+{\tan }^{2}A+co{t}^{2}A$ is an identity.