Question

if, tanA+sinA=a and tanA-sinA then prove that a+b=2sinA.secA​

Answer 1

If tan A + sin A = a …(1)

tan A - sin A = b …(2), how to prove a^2-b^2 = 4(ab)^0.5?

LHS = a^2-b^2 = (a+b)(a-b) = 2tan A*2 sin A

= 4 sin^2 A/ cos A

RHS = 4(ab)^0.5 = 4[(tan A + sin A)(tan A - sin A)]^0.5

= 4[tan^2 A - sin^2 A]^0.5

= 4[(sin^2 A/cos^2 A) - sin^2 A]^0.5

= 4[sin ^2 A - sin^2 A*cos^2 A]^0.5/ cos A

= 4 (sin A/cos A)[1-cos^2]^0.5

= 4 tan A.sin A

= 4 sin^2 A/cos A. Same as LHS, hence

a^2-b^2 = 4(ab)^0.5. Proved

hope it will be useful

please follow me and mark mi as brainleast

Answer 2

Answer:

See the attachment.

Step-by-step explanation: