Question

1+cos Ф/1- cos totally =(cosec+cot tita){2}

Answer 1

❏ Question:-

Prove That,

$\sf\longrightarrow\bf [\frac{1+\cos\theta}{1-\cos\theta}]=(\cosec\theta+\cot\theta)^2$

❏ Solution:-

$\bf L.H.S.\sf\bf= [\frac{1+\cos\theta}{1-\cos\theta}]$

$\sf\bf = [\frac{(1+\cos\theta)\times(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}]$

$\sf\bf = [\frac{(1+\cos\theta)^2}{1^2-\cos^2\theta}]$

$\sf\bf = [\frac{(1+\cos\theta)^2}{\sin^2\theta}]$

$\sf\bf = [\frac{(1+\cos\theta)}{\sin\theta}]^2$

$\sf\bf = [\frac{1}{\sin\theta}+\frac{cos\theta}{\sin\theta}]^2$

$\sf\bf = [\cosec\theta+\cot\theta]^2=R.H.S.\:\:(proved)$

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✰❏ Formula Used:-

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•$\sf\bf\longrightarrow \sin^2\theta=1-\cos^2\theta$

•$\sf\bf\longrightarrow \cosec^2\theta=1+\cot^2\theta$

•$\sf\bf\longrightarrow \sec^2\theta=1+\tan^2\theta$

•$\sf\bf\longrightarrow \cosec\theta=\frac{1}{\sin\theta}$

•$\sf\bf\longrightarrow \sec\theta=\frac{1}{\cos\theta}$

•$\sf\bf\longrightarrow \cot\theta=\frac{1}{\tan\theta}$

•$\sf\bf\longrightarrow (a+b)(a-b)=a^2-b^2$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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