Prove - cos- - -1 - - -1 - -

Given, L.H.S = #160; θ +cos #160; ecθ 1 #160; θ cos #160; ecθ +1 #160; = #160; θ +cos #160; ecθ ( cos #160; e c^ 2 θ ...

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Solving a Quadratic Equation by Completion of Squares Method

Prove : cosθ...

Question

Prove :

$\frac{cos θ + csc θ - 1}{cot θ - csc θ + 1} = csc θ + cot θ$

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\sqrt{\frac{1 + cos θ}{1 - cos θ}} = csc θ + cot θ; 0 \leq θ \leq 90^{o} .

\frac{csc θ}{1 + sec θ} + \frac{1 + sec θ}{csc θ} = \frac{sec θ csc θ + sin θ cos θ + 2 sin θ}{1 + cos θ}

\frac{sin θ}{1 - cos θ} = csc θ + cot θ

Prove the following trigonometric identities:

Prove that:

(i) $\sqrt{\frac{s e c θ - 1}{s e c θ + 1}} + \sqrt{\frac{s e c θ + 1}{s e c θ - 1}} = 2 c o s e c θ$

(ii) $\sqrt{\frac{1 + s i n θ}{1 - s i n θ}} + \sqrt{\frac{1 - s i n θ}{1 + s i n θ}} = 2 c o s e c θ$

(iii) $\sqrt{\frac{1 + c o s θ}{1 - c o s θ}} + \sqrt{\frac{1 - c o s θ}{1 + c o s θ}} = 2 c o s e c θ$

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

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

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