Solving LHS, 1 cos2θ1+cosθ = gt; #160; 2sin^2θ2cos^2θ = gt; #160; tan^2θ #160; = RHSHence proved ...

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Composite Function

1-cos2θ1+cos2...

Question

$\frac{1 - cos 2 θ}{1 + cos 2 θ} = {tan}^{2} θ$

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Solution

Solving LHS,

$\frac{1 - c o s 2 θ}{1 + c o s θ}$

=> $\frac{2 s i n^{2} θ}{2 c o s^{2} θ}$

=> $t a n^{2} θ$ = RHS

Hence proved

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Similar questions

\frac{s i n 2 θ}{1 - c o s 2 θ} = c o t θ o r \frac{1 - c o s^{2} θ}{1 + c o s 2 θ} = \frac{t a n^{2} θ}{2}

\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} = (\cos^{2} θ - \sin^{2} θ)

\frac{1 - \tan^{2} θ}{\cot^{2} θ - 1} = \tan^{2} θ

t a n^{2} θ - \frac{1}{c o s^{2} θ} = - 1

\sqrt{1 + t a n^{2} θ} \sqrt{1 + c o t^{2} θ} \sqrt{1 - c o s^{2} θ} \sqrt{1 - s i n^{2} θ} =

\frac{(1 - {cos}^{2} θ)}{cos θ} \frac{(1 + {tan}^{2} θ)}{tan θ} = tan θ sec θ

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