Question

If
$x=\sqrt{a}tan\alpha$
and
$y=\sqrt{a}sec\alpha$.

Find the value of $y^2–x^2$

• A. a
• B. $\alpha$
• C. a + $\alpha$
• D. a$\alpha$

Answer 1

The correct option is A a

$x=\sqrt{a}tan\alpha$
and
$y=\sqrt{a}sec\alpha$

$x^2=ata{n}^2\alpha$
and
$y^2=ase{c}^2\alpha$

$\Rightarrow y^2-x^2=a.se{c}^2\alpha -a.ta{n}^2\alpha )=a.(se{c}^2\alpha -ta{n}^2\alpha )=a.(1\left)\right[\because 1+ta{n}^2\alpha =se{c}^2\alpha ]=a$