BeMath-Given-tan-x-sec-x-then-sin-x-

Question Number 107900 by bemath last updated on 13/Aug/20

\frac{Σ B e M a t h Σ}{}

G i v e n \tan x - \sec x = ϑ

t h e n \sin x = ?

Answered by bemath last updated on 13/Aug/20

\frac{\sin x - 1}{\cos x} = υ \Rightarrow \sin x - 1 = υ \cos x \dots (1)

\sin x + 1 = - \frac{1}{υ} \cos x \dots (2)

s u b s t i t u t e e q (2) t o e q (1)

\sin x + 1 = - \frac{1}{υ} (\frac{\sin x - 1}{υ})

υ^{2} \sin x + υ^{2} = - \sin x + 1

(υ^{2} + 1) \sin x = 1 - υ^{2}

\Rightarrow \sin x = \frac{1 - υ^{2}}{1 + υ^{2}}

Answered by bobhans last updated on 13/Aug/20

\frac{\land B obhans \land}{}

\tan x - \sec x \times (\frac{\tan x + \sec x}{\tan x + \sec x}) = υ

\frac{\tan^{2} x - \sec^{2} x}{\tan x + \sec x} = υ \to \frac{\tan^{2} x - (1 + \tan^{2} x)}{\tan x + \sec c} = υ

\tan x + \sec x = - \frac{1}{υ} \Rightarrow \frac{\sin x + 1}{\cos x} = - \frac{1}{υ}

\sin x + 1 = - \frac{1}{υ} \cos x

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