Question

tan^2a-tan^2b=sin^2a-sin^2b/cos^2a*cos^b

Answer 1

$tan^{2}$ a - $tan^{2}$ b = $sin^{2}$ a - $sin^{2}$ b / $cos^{2}$ a * $cos^{2}$ b

Solution : As, tan a = sin a / cos a

  so, $tan^{2}$ a = $sin^{2}$ a / $cos^{2}$

  ∴LHS = $sin^{2}$a / $cos^{2}$ a - $sin^{2}$ b / $cos^{2}$ b

  By taking LCM,

$sin^{2}$ a * $cos^{2}$ b - $sin^{2}$ b * $cos^{2}$ a / $cos^{2}$ a $cos^{2}$ b

= $sin^{2} a - sin^{2} b$ / $cos^{2} a - cos^{2} b$ = RHS

Hence proved

I hope it helps you

Thank you