Question

Prove that ${(sinA+cosecA)}^2+{(cosA+secA)}^2=7+{tan}^{2}A+{cot}^{2}A$

Answer 1

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

Taking LHS

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

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

$[\because 2\sin A\cdot cosecA=2$ and $2\cos A\cdot \sec A=2]$

$={\sin }^{2}A+{\cos }^{2}A+1/{\sin }^{2}A+1/{\cos }^{2}A+4$

$=1/{\sin }^{2}A+1/{\cos }^{2}A+5$ $[\because {\sin }^{2}A+{\cos }^{2}A=1]$

$=({\sin }^{2}A+{\cos }^{2}A+5{\sin }^{2}A{\cos }^{2}A)/{\sin }^{2}A\cdot {\cos }^{2}A$

$=(1+5{\sin }^{2}A\cdot {\cos }^{2}A)/{\sin }^{2}A\cdot {\cos }^{2}A$

$=({\sec }^{2}A\cdot cose{c}^{2}A)+5$ ………..$\left(1\right)$

Taking RHS

$={\tan }^{2}A+{\cot }^{2}A+7$

$=\frac{{\sin }^{2}A}{{\cos }^{2}A}+\frac{{\cos }^{2}A}{{\sin }^{2}A}+7$

$=\frac{({\sin }^{2}A+{\cos }^{2}A{)}^2-2{\sin }^{2}A\cdot {\cos }^{2}A}{{\sin }^{2}A\cdot {\cos }^{2}A}+7$

$=(cose{c}^{2}A.{\sec }^{2}A)-2+7$

$=(cose{c}^{2}A\cdot {\sec }^{2}A)+5$ ……………….$\left(2\right)$

$\therefore$ LHS$=$RHS

Hence proved.