Question

cotA+tanA/cotA-tanA=sec2A

Answer 1

Here,

LHS = (cotA + tanA) ÷ (cotA - tanA)

= (cosA/sinA + sinA/cosA) ÷ (cosA/sinA - sinA/cosA)

= (cos^2 A + sin^2 A) ÷ (cos^2 A - sin^2 A)

By trigonometric identities,

sin^2 A + cos^2 A = 1 and cos^2 A - sin^2 A = cos2A

So LHS will become,

= (1) ÷ cos 2A

= sec 2A

=RHS

Please mark my answer as the brainliest.

[THANK YOU]