Question

Given, A+B=90
Prove that
$\sqrt{ \frac{ \tan \alpha \tan \beta + \tan \alpha \cot \beta }{ \sin\alpha \sec \beta } - \frac{ { \sin( \beta ) }^{2} }{ \cos( { \beta }^{2} ) } } = \tan( \alpha )$

​

Answer 1

Answer:

Step-by-step explanation:

Given, LHS = root tan a tan b + tan a cot b/sin a sec b - sin^2 b/cos^2 a

      = root tan a tan(90-a) + tan a cot(90-a)/sin a sec(90-a) - sin^2(90-a)/cos^2 a

      = root tan a.cot a + tan a.tana/sin a.cosec a - cos^2 a/cos^2 a

      = root 1+tan^2 a/1 - 1

     = root tan^2 a

    = tan a.

Mark as brainliest please