Question

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

Answer 1

Now,

Considering $LHS=(\sin \alpha +\csc \alpha {)}^2+(\cos \alpha +\sec \alpha {)}^2$

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

$=({\sin }^2\alpha +{\cos }^2\alpha )+({\csc }^2\alpha +{\sec }^2\alpha )+4$

$=1+(1+{\cot }^2\alpha +1+{\tan }^2\alpha )+4$ [ Since ${\csc }^2\alpha -{\cot }^2\alpha =1$ and ${\sec }^2\alpha -{\tan }^2\alpha =1$]

$=7+{\tan }^2\alpha +{\cot }^2\alpha =RHS$

Hence, proved.