Question

If ${\tan }^{2}y$ $co{\sec }^{2}x-1={\tan }^{2}y$, then which one of the following is correct?

• A. $x-y=0$
• B. $x=2y$
• C. $y=2x$
• D. $x-y=1^o$

Answer 1

The correct option is A $x-y=0$

$ta{n}^{2}ycose{c}^{2}x-1=ta{n}^{2}y\Rightarrow ta{n}^{2}ycose{c}^{2}x=se{c}^{2}y$

$\Rightarrow si{n}^{2}y=si{n}^{2}x\Rightarrow x=y\&x-y=0$

(use $ta{n}^2=se{c}^{2}y-1)$