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प्रश्न
Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ
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उत्तर
L.H.S = `(tan^2 theta - 1)/(tan^2 theta + 1)`
= `(tan^2 theta - 1)/(sec^2 theta)`
= `(sin^2 theta)/(cos^2 theta) - 1 ÷ 1/(cos^2 theta)`
= `(sin^2 theta - cos^2 theta)/(cos^2 theta) xx (cos^2 theta)/1`
= sin2θ − cos2θ
= 1 − cos2θ − cos2θ
= 1 – cos2θ
L.H.S = R.H.S
Hence it is proved.
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