Question

Solve:
$(\sin A+\csc A{)}^2+(\cos A+\sec A{)}^2=7+{\tan }^{2}A+{\cot }^{2}A$

Answer 1

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

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

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

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

$=5+\frac{{\sin }^{2}A+{\cos }^{2}A}{{\sin }^{2}A{\cos }^{2}A}$

$=5+\frac{1}{{\sin }^{2}A{\cos }^{2}A}$ since ${\sin }^{2}A+{\cos }^{2}A=1$

$=5+{\sec }^{2}A{\csc }^{2}A$

$=5+(1+{\tan }^{2}A)(1+{\cot }^{2}A)$

$=5+1+{\cot }^{2}A+{\tan }^{2}A+{\tan }^{2}A{\cot }^{2}A$

$=6+{\cot }^{2}A+{\tan }^{2}A+1$ since $\tan A\cot A=1$

$=7+{\cot }^{2}A+{\tan }^{2}A=$R.H.S

Hence proved.