Prove that - -1-cos2-1-cos 2-tan-

Solving LHS. we get √(1 cos2θ1+cos2θ) √(2sin^2θ2cos^2θ) √(sin^2θcos^2θ) √(tan^2θ) tanθ = RHSHence proved. ...

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Principal Solution of Trigonometric Equation

Prove that : ...

Question

Prove that :
$\sqrt{\frac{1 - cos 2 θ}{1 + cos 2 θ}} = tan θ$

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

x = \sqrt{{cos}^{2} θ + \sqrt{{cos}^{2} θ {\sqrt{cos}}^{2} θ + . . . \infty}}

y = \sqrt{sec θ - \sqrt{sec θ - \sqrt{sec θ - . . . \infty}}}

(x^{2} - x) (y^{2} + y) = 1

z = \sqrt{tan θ + \sqrt{tan θ + \sqrt{tan θ + . . . \infty}}}

z^{2} = tan θ + z

Prove that :

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

c o s e c θ \sqrt{1 - {cos}^{2} θ} = 1

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