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Question Number 35654 by chakraborty ankit last updated on 21/May/18

$$if\cos {}^2\theta -\sin {}^2\theta =\tan {}^2\mathrm{\varnothing }Thenproofthat$$$$2\cos {}^2\mathrm{\varnothing }-1=\cos {}^2\mathrm{\varnothing }-\sin {}^2\mathrm{\varnothing }=2\tan {}^2\theta$$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/May/18

$${cos}^2\theta -{sin}^2\theta ={tan}^2\mathrm{\varnothing }thatmeans$$$$cos2\theta ={tan}^2\mathrm{\varnothing }$$$$2{cos}^2\mathrm{\varnothing }-1$$$$=cos2\varphi$$$$=\frac{1-{tan}^2\mathrm{\varnothing }}{1+{tan}^2\mathrm{\varnothing }}$$$$=\frac{1-cos2\theta }{1+cos2\theta }$$$$=\frac{2{sin}^2\theta }{2{cos}^2\theta }$$$$={tan}^2\theta$$$$plschec\stackrel{}{k}$$

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