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Question Number 116630 by bobhans last updated on 05/Oct/20

Given cosec x + \cot x = p, find the

value of cosec x = ?

Answered by TANMAY PANACEA last updated on 05/Oct/20

1 + {c o t}^{2} x = {c o s e c}^{2} x

1 = (c o s e c x + c o t x) (c o s e c x - c o t x)

c o s e c x + c o t x = p

c o s e c x - c o t x = \frac{1}{p}

2 c o s e c x = (p + \frac{1}{p})

c o s e c x = \frac{1}{2} (p + \frac{1}{p})

Answered by MJS_new last updated on 05/Oct/20

\frac{1}{\sin x} + \frac{\cos x}{\sin x} = p

\frac{1 + \sqrt{1 - \sin^{2} x}}{\sin x} = p

\Rightarrow \sin x = \frac{2 p}{p^{2} + 1} \Rightarrow cosec x = \frac{p^{2} + 1}{2 p}

Answered by bemath last updated on 05/Oct/20

\frac{1}{\sin x} + \frac{\cos x}{\sin x} = p \Rightarrow 1 + \cos x = psin x

\Rightarrow \sqrt{1 - \sin^{2} x} = psin x - 1

\Rightarrow 1 - \sin^{2} x = p^{2} \sin^{2} x - 2 psin x + 1

(1 + p^{2}) \sin^{2} x = 2 psin x

\sin x \neq 0 \Rightarrow (1 + p^{2}) \sin x = 2 p

\Rightarrow \sin x = \frac{2 p}{1 + p^{2}} and cosec x = \frac{1 + p^{2}}{2 p}

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