Question

Prove that
$(\sin A+\csc A{)}^2+(\cos A+\sec A{)}^2=6+{\csc }^{2}A{\sec }^{2}A$

Answer 1

L.H.S$={(\sin A+\csc A)}^2+{(\cos A+\sec A)}^2$

$={\sin }^{2}A+{\csc }^{2}A+2\sin A\csc A+{\cos }^{2}A+{\sec }^{2}A+2\cos A\sec A$

$=({\sin }^{2}A+{\cos }^{2}A)+({\csc }^{2}A+{\sec }^{2}A)+2\sin A\csc A+2\cos A\sec A$

$=1+({\csc }^{2}A+{\sec }^{2}A)+2\sin A\csc A+2\cos A\sec A$ since ${\sin }^2\theta +{\cos }^2\theta =1$

$=2+(\frac{1}{{\sin }^{2}A}+\frac{1}{{\cos }^{2}A})+2\sin A\times \frac{1}{\sin A}+2\cos A\times \frac{1}{\cos A}$ since $\csc A=\frac{1}{\sin A}$ and $\sec A=\frac{1}{\cos A}$

$=2+(\frac{1}{{\sin }^{2}A}+\frac{1}{{\cos }^{2}A})+2+2$

$=6+\frac{{\sin }^{2}A+{\cos }^{2}A}{{\sin }^{2}A{\cos }^{2}A}$

$=6+\frac{1}{{\sin }^{2}A{\cos }^{2}A}$

$=6+{\csc }^{2}A{\sec }^{2}A=$R.H.S

Hence proved.