Question

Prove that sin theta +cos theta /sin theta - cos theta +sin theta - cos theta /sin theta +cos theta =2sec²theta /tan²theta - 1

Answer 1

follow the above steps

Answer 2

TO PROVE :–

\begin{gathered} \\ \bf \implies \dfrac{\sin \theta + \cos \theta}{ \sin \theta - \cos\theta} + \dfrac{\sin \theta - \cos\theta}{ \sin \theta + \cos\theta}= \dfrac{2 \sec ^{2} \theta}{ \tan^{2} \theta - 1} \\ \end{gathered}

⟹

sinθ−cosθ

sinθ+cosθ

+

sinθ+cosθ

sinθ−cosθ

=

tan

2

θ−1

2sec

2

θ

SOLUTION :–

• Let's take L.H.S. –

\begin{gathered} \\ \bf \: \: = \: \: \dfrac{\sin \theta + \cos \theta}{ \sin \theta - \cos\theta} + \dfrac{\sin \theta - \cos\theta}{ \sin \theta + \cos\theta}\\ \end{gathered}

=

sinθ−cosθ

sinθ+cosθ

+

sinθ+cosθ

sinθ−cosθ

\begin{gathered} \\ \bf \: \: = \: \: \dfrac{(\sin \theta + \cos \theta)^{2} +(\sin \theta - \cos\theta)^{2} }{(\sin \theta - \cos\theta)(\sin \theta + \cos\theta)} \\ \end{gathered}

=

(sinθ−cosθ)(sinθ+cosθ)

(sinθ+cosθ)

2

+(sinθ−cosθ)

2

\begin{gathered} \\ \bf \: \: = \: \: \dfrac{(\sin^{2} \theta + \cos^{2} \theta + 2\sin \theta \cos\theta) +(\sin^{2} \theta + \cos^{2} \theta - 2\sin \theta \cos\theta)}{(\sin \theta - \cos\theta)(\sin \theta + \cos\theta)} \\ \end{gathered}

=

(sinθ−cosθ)(sinθ+cosθ)

(sin

2

θ+cos

2

θ+2sinθcosθ)+(sin

2

θ+cos

2

θ−2sinθcosθ)

=

(sin

2

θ−cos

2

θ)

(sin

2

θ+cos

2

θ+2sinθcosθ)+(sin

2

hence proved