Question

tan theeta by 1-cot theeta +cot theeta by 1-tan theeta =1+sec theeta cosec theeta​

Answer 1

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Answer 2

$\huge\sf\blue{Given}$

✭ $\sf \dfrac{tan\theta}{1-cot\theta}+\dfrac{co\theta}{1-tan\theta}=1+sec\theta\times cosec\theta$

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$\huge\sf\gray{To\;Prove}$

◈ The above expression

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$\huge\sf\purple{Steps}$

Convert this into- sinθ and cosθ

$\bigstar\sf \ tan\theta= \dfrac{sin\theta}{cos\theta}\\ \\ \bigstar \ \sf cot\theta= \dfrac{cos\theta}{sin\theta}$

Substituting values,

In LHS :

$\dashrightarrow\sf \dfrac{tan\theta}{1-cot\theta}+\dfrac{cot\theta}{1-tan\theta}\\ \\ \dashrightarrow\sf \dfrac{\bigg(\dfrac{sin\theta}{cos\theta}\bigg)}{\bigg(1-\dfrac{cos\theta}{sin\theta}\bigg)}\ + \dfrac{\bigg(\dfrac{cos\theta}{sin\theta}\bigg)}{\bigg(1-\dfrac{sin\theta}{cos\theta}\bigg)}\\ \\ \dashrightarrow\sf \dfrac{\bigg(\dfrac{sin\theta}{cos\theta}\bigg)}{\bigg(\dfrac{sin\theta-cos\theta}{sin\theta}\bigg)} \ + \dfrac{\bigg(\dfrac{cos\theta}{sin\theta}\bigg)}{\bigg(\dfrac{cos\theta-sin\theta}{cos\theta}\bigg)} \\ \\ \dashrightarrow\sf \bigg(\dfrac{sin\theta}{cos\theta}\bigg)\times \bigg(\dfrac{sin\theta}{sin\theta-cos\theta}\bigg)\ + \bigg(\dfrac{cos\theta}{sin\theta}\bigg)\times \bigg(\dfrac{cos\theta}{cos\theta-sin\theta}\bigg)\\ \\ \dashrightarrow\sf \dfrac{sin^2\theta}{cos\theta(sin\theta - cos\theta)} \ + \dfrac{cos^2\theta}{sin\theta(cos\theta-sin\theta)}$

$\sf Take (-) as common$

$\twoheadrightarrow\sf \dfrac{sin^2\theta}{cos\theta(sin\theta - cos\theta)} \ - \dfrac{cos^2\theta}{sin\theta(sin\theta-cos\theta)}\\ \\ \twoheadrightarrow\sf \dfrac{sin^3\theta-cos^3\theta}{sin\theta\ cos\theta(sin\theta-cos\theta)}$

We use the formula,

$\bullet{\sf {\small\ \ a^3-b^3= (a-b)(a^2+b^2+ab)}}$

$\leadsto\sf \dfrac{\cancel{(sin\theta-cos\theta)}(sin^2\theta+cos^2\theta+sin\theta. cos\theta)}{sin\theta cos\theta\cancel{(sin\theta-cos\theta)}}$

$\bullet{\sf {\small\ \ sin^2\theta+cos^2\theta=1}}$

$\dashrightarrow\sf \dfrac{1+ sin\theta \ cos\theta}{sin\theta\ cos\theta}\\ \\ \dashrightarrow\sf \dfrac{1}{sin\theta \ cos\theta} + \cancel{\dfrac{sin\theta.cos\theta}{sin\theta\ cos\theta}}\\ \\ \bullet\sf \dfrac{1}{sin\theta} = cosec \theta\ \\ \bullet\sf \dfrac{1}{cos\theta}= sec\theta\\ \\ \dashrightarrow\sf cosec\theta sec\theta+1\\ \\ \dashrightarrow\sf \ or \ \ 1+ cosec\theta sec\theta\: \: (\ Hence \ Proved!!)\\ \\ \sf \ L.H.S = R.H.S$

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