Question

1-sectheta+tantheta/1+sectheta-tantheta= sectheta + tantheta-1/sectheta+tantheta+1​

Answer 1

$\textbf{To prove:}$

$\mathsf{\dfrac{1-sec\theta+tan\theta}{1+sec\theta-tan\theta}=\dfrac{sec\theta+tan\theta-1}{sec\theta+tan\theta+1}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{\dfrac{1-sec\theta+tan\theta}{1+sec\theta-tan\theta}}$

$\mathsf{Using,}\;\boxed{sec^2A-tan^2A=1}$

$\mathsf{=\dfrac{(sec^2\theta-tan^2\theta)-sec\theta+tan\theta}{(sec^2\theta-tan^2\theta)+sec\theta-tan\theta}}$

$\mathsf{=\dfrac{(sec\theta-tan\theta)(sec\theta+tan\theta)-(sec\theta-tan\theta)}{(sec\theta-tan\theta)(sec\theta+tan\theta)+(sec\theta-tan\theta)}}$

$\mathsf{=\dfrac{(sec\theta-tan\theta)[sec\theta+tan\theta-1]}{(sec\theta-tan\theta)[sec\theta+tan\theta+1]}}$..

$\mathsf{=\dfrac{sec\theta+tan\theta-1}{sec\theta+tan\theta+1}}$

$\implies\boxed{\mathsf{\dfrac{1-sec\theta+tan\theta}{1+sec\theta-tan\theta}=\dfrac{sec\theta+tan\theta-1}{sec\theta+tan\theta+1}}}$

$\textbf{Find more:}$

If cos2x-cosx= sin4x - sinx(

where tan doesnot equal to 1) find value of cos3x - sin3x

https://brainly.in/question/19348427

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

https://brainly.in/question/4927186

Answer 2

$\textbf{To prove:}$

$\mathsf{\dfrac{1-sec\theta+tan\theta}{1+sec\theta-tan\theta}=\dfrac{sec\theta+tan\theta-1}{sec\theta+tan\theta+1}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{\dfrac{1-sec\theta+tan\theta}{1+sec\theta-tan\theta}}$

$\mathsf{Using,}\;\boxed{sec^2A-tan^2A=1}$

$\mathsf{=\dfrac{(sec^2\theta-tan^2\theta)-sec\theta+tan\theta}{(sec^2\theta-tan^2\theta)+sec\theta-tan\theta}}$

$\mathsf{=\dfrac{(sec\theta-tan\theta)(sec\theta+tan\theta)-(sec\theta-tan\theta)}{(sec\theta-tan\theta)(sec\theta+tan\theta)+(sec\theta-tan\theta)}}$

$\mathsf{=\dfrac{(sec\theta-tan\theta)[sec\theta+tan\theta-1]}{(sec\theta-tan\theta)[sec\theta+tan\theta+1]}}$..

$\mathsf{=\dfrac{sec\theta+tan\theta-1}{sec\theta+tan\theta+1}}$

$\implies\boxed{\mathsf\red{\dfrac{1-sec\theta+tan\theta}{1+sec\theta-tan\theta}=\dfrac{sec\theta+tan\theta-1}{sec\theta+tan\theta+1}}}$