Question

$\frac{{\tan }^2\alpha -{\sin }^2\alpha }{{\cot }^2\alpha -{\cos }^2\alpha }=(\tan \alpha {)}^K$ ,where $K\in N$, the value of $K$ is equal to -

• A. $8$
• B. $6$
• C. $4$
• D. $2$

Answer 1

The correct option is B $6$

$\frac{{\tan }^2\alpha -{\sin }^2\alpha }{{\cot }^2\alpha -{\cos }^2\alpha }=\frac{{\tan }^2\alpha -{\tan }^2\alpha {\cos }^2\alpha }{\frac{1}{{\tan }^2\alpha }-\frac{{\sin }^2\alpha }{{\tan }^2\alpha }}$

$=\frac{{\tan }^2\alpha {\tan }^2\alpha (1-{\cos }^2\alpha )}{1-{\sin }^2\alpha }$

$={\tan }^2\alpha {\tan }^2\alpha ({\sec }^2\alpha -1)={\tan }^6\alpha$

On comparing it with $(\tan \alpha {)}^K$, we get value of $K$ as $6$.