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प्रश्न
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
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उत्तर
We have , `(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) `
=`((sin theta + cos theta )^2 + (sin theta - cos theta)^2) /((sin theta - cos theta )(sin theta + cos theta))`
=`(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)`
=`(1+1)/(sin^2 theta - cos^2 theta)`
=`2/(sin^2 theta - cos^2 theta)`
Again ,` 2/(sin^2 theta - cos^2 theta)`
=`2/(sin^2 theta -(1-sin^2 theta))`
=`2/(2 sin ^2 theta -1)`
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= `1/(sinθ xx cosθ)` ....... ∵ `square`
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∴ L.H.S. = R.H.S.
