Question

Tan theeta/1-cot there + cot theeta/ 1-tan theeta = 1+sec theeta . Cosec theeta

Answer 1

Given :-

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

We have to prove the given equation :-

Convert this into- $\sf sin\theta\ \ and \ cos\theta$

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

Put the value of , in the given equation

Take LHS :-

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

Take (-) as common ,

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

Formula used here , !

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

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

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

$\longmapsto\sf \dfrac{1+ sin\theta \ cos\theta}{sin\theta\ cos\theta}\\ \\ \\ \longmapsto\sf \dfrac{1}{sin\theta \ cos\theta} + \cancel{\dfrac{sin\theta.cos\theta}{sin\theta\ cos\theta}}\\ \\ \\ \bullet\sf \dfrac{1}{sin\theta} = cosec \theta\ ; \ \bullet\sf \dfrac{1}{cos\theta}= sec\theta\\ \\ \\ \longmapsto\sf cosec\theta sec\theta+1\\ \\ \\ \longmapsto\sf \ or \ \ 1+ cosec\theta sec\theta\ \ \ \ (\ hence \ proved\ !)\\ \\ \sf \ L.H.S = R.H.S$

Answer 2

${\underline{\huge{\mathbf{\color{blue}{question}}}}}$

$\sf \dfrac{tan\theta}{1-cot\theta}+\dfrac{cot\theta}{1-tan\theta}=1+sec\theta. cosec\theta$

${\underline{\huge{\mathbf{\color{blue}{solution}}}}}$

$\sf \dfrac{tan\theta}{1-cot\theta}+\dfrac{cot\theta}{1-tan\theta}=1+sec\theta. cosec\theta$

LHS :-

$\sf \dfrac{tan\theta}{1-cot\theta}+\dfrac{cot\theta}{1-tan\theta}$

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

$\sf \dfrac{\bigg(\dfrac{sin\theta}{cos\theta}\bigg)}{\bigg(1-\dfrac{cos\theta}{sin\theta}\bigg)}\ + \dfrac{\bigg(\dfrac{cos\theta}{sin\theta}\bigg)}{\bigg(1-\dfrac{sin\theta}{cos\theta}\bigg)}$

$\sf \dfrac{\bigg(\dfrac{sin\theta}{cos\theta}\bigg)}{\bigg(\dfrac{sin\theta-cos\theta}{sin\theta}\bigg)} \ + \dfrac{\bigg(\dfrac{cos\theta}{sin\theta}\bigg)}{\bigg(\dfrac{cos\theta-sin\theta}{cos\theta}\bigg)}$

$\sf \bigg(\dfrac{sin\theta}{cos\theta}\bigg)\times \bigg(\dfrac{sin\theta}{sin\theta-cos\theta}\bigg)\ + \bigg(\dfrac{cos\theta}{sin\theta}\bigg)\times \bigg(\dfrac{cos\theta}{cos\theta-sin\theta}\bigg)$

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

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

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

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

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

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

$\sf \dfrac{1+ sin\theta \ cos\theta}{sin\theta\ cos\theta}$

$\sf \dfrac{1}{sin\theta \ cos\theta} + \cancel{\dfrac{sin\theta.cos\theta}{sin\theta\ cos\theta}}$

• $\sf \dfrac{1}{sin\theta} = cosec \theta$
• $\sf \dfrac{1}{cos\theta}= sec\theta$

$cosec\theta \:sec\theta + 1$

RHS:-

$cosec\theta\: sec\theta + 1$

compare

LHS = RHS

......hence proved