Question

If ${\cos }^2\alpha -{\sin }^2\alpha ={\tan }^2\beta ,$ then show that ${\tan }^2\alpha ={\cos }^2\beta -{\sin }^2\beta$.

Answer 1

As we know that, $1+{\tan }^2\theta ={\sec }^2\theta$

So, we can write

${\cos }^2\alpha -{\sin }^2\alpha ={\sec }^2\beta -1\Rightarrow {\sec }^2\beta ={\cos }^2\alpha -{\sin }^2\alpha +1\Rightarrow {\cos }^2\beta =\frac{1}{{\cos }^2\alpha -{\sin }^2\alpha +({\sin }^2\alpha +{\cos }^2\alpha )}(as{\sin }^2\alpha +{\cos }^2\alpha =1)\Rightarrow {\cos }^2\beta =\frac{1}{2{\cos }^2\alpha }$

also, ${\sin }^2\beta =1-{\cos }^2\beta =1-\frac{1}{2{\cos }^2\alpha }$

So, ${\cos }^2\beta -{\sin }^2\beta =\frac{1}{2{\cos }^2\alpha }-(1-\frac{1}{2{\cos }^2\alpha })$

$=\frac{1}{2{\cos }^2\alpha }-1+\frac{1}{2{\cos }^2\alpha }=\frac{1}{{\cos }^2\alpha }-1={\sec }^2\alpha -1={\tan }^2\alpha$