Question

If $(\sin \alpha +\text{cosec}\alpha {)}^2+(\cos \alpha +\sec \alpha {)}^2=k^2+{\tan }^2\alpha +{\cot }^2\alpha$, then $k^2=$

• A. $9$
• B. $7$
• C. $5$
• D. $3$

Answer 1

Answer: (B)

$7$

We need to prove $(\sin \alpha +\text{cosec}\alpha {)}^2+(\cos \alpha +\sec \alpha {)}^2=k^2+{\tan }^2\alpha +{\cot }^2\alpha$

Taking LHS $={\sin }^2\alpha +{\text{cosec}}^2\alpha +2+{\cos }^2\alpha +{\sec }^2\alpha +2$

$=5+1+{\tan }^2\alpha +1+{\cot }^2\alpha$
$=7+{\cot }^2\alpha +{\tan }^2\alpha$
Therefore, $k^2=7$