Question

$\huge \fbox \red{❥ Question}$
Prove that sin theta/1-cos theta + tan theta /1 + cos theta = sec theta * cosec theta + cot theta?​

Answer 1

Answer:

Sinθ/(1 – cosθ) + Tanθ/(1 + cosθ) = Secθ.Cosecθ + Cotθ

Let us start with LHS

= Sinθ/(1 – cosθ) + Tanθ/(1 + cosθ)

= (sinθ(1 + cosθ) + Tanθ(1-Cosθ))/(1 – Cos²θ)

= (sinθ(1 + cosθ) + (Tanθ – Sinθ)) /Sin²θ

= ( 1 + cosθ + 1/Cosθ – 1)/Sinθ

= (cosθ + 1/Cosθ)/Sinθ

= 1/CosθSinθ + cosθ/Sinθ

= Secθ.Cosecθ + Cotθ

= RHS

Hence Proved

Answer 2

${\bf{To~Prove~:~\sf{\dfrac{sin \theta}{1 - cos \theta}}~ +~ {\dfrac{tan \theta}{1 + cos \theta}} ~= ~cot \theta ~+~ sec \theta cosec \theta}}$

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◗L.H.S. = ${\sf{ \dfrac{sin \theta}{1 - cos \theta} + \dfrac{tan \theta}{1 + cos \theta}}}$

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${\sf:\implies{\sf{ \dfrac{ sin \theta (1 + cos \theta ) + tan \theta (1 - cos \theta )}{(1 + cos \theta)(1 - cos \theta)}}}}$

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• ${\boxed{\sf\purple{Identity \ : \ (a + b)(a - b) = a^2 - b^2}}}$

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◗${\sf{Here, \ a = 1, \ b = cos \theta}}$

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${\sf:\implies{\dfrac{ sin \theta (1 + cos \theta ) + tan \theta (1 - cos \theta )}{(1)^2 - (cos \theta)^2}}}$

${\sf:\implies{\dfrac{ sin \theta (1 + cos \theta ) + tan \theta (1 - cos \theta )}{1 - cos^2 \theta}}}$

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• ${\boxed{\sf\purple{Identity \ : \ sin^2 \theta + cos^2 \theta = 1}}}$

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◗${\sf{From \ this, \ we \ get \ [ 1 - cos^2 \theta = sin^2 \theta ]}}$

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${\sf:\implies{\dfrac{ sin \theta (1 + cos \theta ) + tan \theta (1 - cos \theta )}{sin^2 \theta}}}$

${\sf:\implies{\dfrac{ sin \theta + sin \theta cos \theta + tan \theta - tan \theta cos \theta}{sin^2 \theta}}}$

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• ${\boxed{\sf\pink{Identity \ : \ tan \theta = {\dfrac{sin \theta}{cos \theta}}}}}$

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${\sf:\implies{\dfrac{ sin \theta + sin \theta cos \theta + tan \theta -~sin \theta}{cos \theta} \times cos \theta~sin^2 \theta}}$

${\sf:\implies{\dfrac{ sin \theta + sin \theta cos \theta + tan \theta - sin \theta}{sin^2 \theta}}}$

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◗Rearranging the terms in the numerator.

${\sf:\implies{\dfrac{ sin \theta - sin \theta + sin \theta cos \theta + tan \theta}{sin^2 \theta}}}$

${\sf:\implies{\dfrac{sin \theta cos \theta + tan \theta}{sin^2 \theta}}}$

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• ${\boxed{\sf\pink{Identity \ : \ tan \theta = {\dfrac{sin \theta}{cos \theta}}}}}$

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${\sf:\implies{\dfrac{sin \theta cos \theta + sin \theta}{cos \theta ~sin^2 \theta}}}$

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

${\sf:\implies{ \dfrac{sin \theta cos^2 \theta + sin \theta}{cos \theta} sin^2 \theta}}$

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

${\sf:\implies{\dfrac{sin \theta cos^2 \theta + sin \theta }{cos \theta . sin^2 \theta}}}$

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◗Taking common terms out from numerator and denominator.

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${\sf:\implies{{\dfrac{sin \theta (cos^2 \theta + 1) }{sin \theta (cos \theta . sin \theta)}}}}$

${\sf:\implies{\dfrac{cos^2 \theta + 1}{cos \theta . sin \theta}}}$

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◗$\underline{\frak{We ~can~ write~ this ~as~ :}}$

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${\sf:\implies{\dfrac{cos^2 \theta}{cos \theta sin \theta}} + {\dfrac{1}{cos \theta sin \theta}}}$

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

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◗$\underline{\frak{We ~can~ write~ this ~as~ :}}$

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${\sf:\implies {\dfrac{cos \theta}{sin \theta}} + {\dfrac{1}{cos \theta}} \times {\dfrac{1}{sin \theta}}}$

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• ${\boxed{\sf\pink{Identity \ : \ {\dfrac{cos \theta}{sin \theta}} = cot \theta}}}$
• ${\boxed{\sf\pink{Identity \ : \ {\dfrac{1}{cos \theta}} = sec \theta}}}$
• ${\boxed{\sf\pink{Identity \ : \ {\dfrac{1}{sin \theta}}~ = ~cosec \theta}}}$

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${\sf:\implies{ cot \theta + sec \theta cosec \theta + cot \theta}}$= R.H.S.

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Hence, proved !!