Question

sintheta -cos theta /sin theta +cos theta +sin theta +cos theta /sin theta -cos theta =2/2sin²theta -1

Answer 1

Answer:

Step-by-step explanation:

We have to prove that $\frac{(sin\theta-cos\theta)}{(sin\theta+cos\theta)}+\frac{(sin\theta+cos\theta)}{(sin\theta-cos\theta)}=\frac{2}{2sin^{2}\theta-1 }$

We take the left hand side of the given equation.$\frac{(sin\theta-cos\theta)}{(sin\theta+cos\theta)}+\frac{(sin\theta+cos\theta)}{(sin\theta-cos\theta)}=\frac{(sin\theta-cos\theta)(sin\theta-cos\theta)+(sin\theta+cos\theta)(sin\theta+cos\theta)}{(sin\theta+cos\theta)(sin\theta-cos\theta)}$

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

                                     = $\frac{1+1}{sin^{2}\theta-(1-sin^{2}\theta ) }$ [SInce $sin^{2}\theta + cos^{2}\theta = 1$]

                                     = $\frac{2}{2sin^{2}\theta-1}$

                                     = Left hand side of the equation

Hence proved.

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Answer 2

Answer:

LHS=RHS

Step-by-step explanation:

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