Question

How to prove the following:
1+sin2A/1-sin2A=(cotA+1/cotA-1)2

Answer 1

$\underline{\textsf{To prove:}}$

$\mathsf{\dfrac{1+sin2A}{1-sin2A}=(\dfrac{cotA+1}{cotA-1})^2}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\dfrac{1+sin2A}{1-sin2A}}$

$\mathsf{=\dfrac{cos^2A+sin^2A+2\,cosA\,sinA}{cos^2A+sin^2A-2\,cosA\,sinA}}$

$\mathsf{=\dfrac{(cosA+sinA)^2}{(cosA-sinA)^2}}$

$\textsf{Taking sinA common from both numerator and denominator}$

$\mathsf{=\dfrac{sin^2A(\dfrac{cosA}{sinA}+1)^2}{sin^2A(\dfrac{cosA}{sinA}-1)^2}}$

$\mathsf{=\dfrac{(\dfrac{cosA}{sinA}+1)^2}{(\dfrac{cosA}{sinA}-1)^2}}$

$\mathsf{=\dfrac{(cotA+1)^2}{(cotA-1)^2}}$

$\implies\boxed{\mathsf{\dfrac{1+sin2A}{1-sin2A}=(\dfrac{cotA+1}{cotA-1})^2}}$

Find more:

Prove that 1-sin^2x/1+cotx-cos^2x/1+tanx=sinxcossx

https://brainly.in/question/4927186

Tan(π/4+x)-tan(π/4-x)=2tan 2x

https://brainly.in/question/10999329