1-cos 2 -sin 2 -1-cos 2 -sin 2 -tan-

1 cos2θ+sin2θ/1+cos2θ+sin2θ=tanθ=1 cos^2θ+sin^2θ+2sinθ cosθ/1 cos^2θ sin^2θ+2sinθ cosθ =(1 cos^2θ)+sin^2θ+2sinθ cosθ/(1 sin^2θ)+cos^2θ+2sinθ cosθ =2sin^2θ+2sinθ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

1-cos 2 θ+sin...

Question

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

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Solution

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

Suggest Corrections

1−cos2θ+sin2θ1+cos2θ+sin2θ=tanθ

1−cos2θ+sin2θ1+cos2θ+sin2θ=tanθ=1−cos2θ+sin2θ+2sinθcosθ1−cos2θ−sin2θ+2sinθcosθ =(1−cos2θ)+sin2θ+2sinθcosθ(1−sin2θ)+cos2θ+2sinθcosθ =2sin2θ+2sinθcosθ2cos2θ+2sinθcosθ=2sinθ2cosθ(sinθ+cosθcosθ+sinθ)=tanθ

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