Question

sintheta+costheta/sintheta-costheta +sintheta-costheta/sintheta=cos thaeta

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{\dfrac{sin\theta+cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}}$

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

$\mathsf{\dfrac{sin\theta+cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}}$

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

$\underline{\textsf{Formula used:}}$

$\boxed{\begin{minipage}{4cm} \\\mathsf{sin^2A+cos^2A=1}\\\\\mathsf{cos2A=cos^2A-sin^2A}\\ \end{minipage}}$

$\textsf{Consider,}$

$\mathsf{\dfrac{sin\theta+cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}}$

$\mathsf{=\dfrac{(sin\theta+cos\theta)^2+(sin\theta-cos\theta)^2}{(sin\theta-cos\theta)(sin\theta+cos\theta)}}$

$\mathsf{=\dfrac{sin^2\theta+cos^2\theta+2\,sin\theta\,cos\theta+sin^2\theta+cos^2\theta-2\,sin\theta\,cos\theta}{sin^2\theta-cos^2\theta}}$

$\mathsf{=\dfrac{1+1}{sin^2\theta-cos^2\theta}}$

$\mathsf{=\dfrac{2}{-(cos^2\theta-sin^2\theta)}}$

$\mathsf{=\dfrac{2}{-cos\,2\theta}}$

$\mathsf{=-2\,sec\,2\theta}$

$\underline{\textsf{Answer:}}$

$\textsf{The simplified form of the given expression is}\;\mathsf{-2\,sec2\theta}$

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