Question

prove that tantheta÷1-cottheta+cottheta÷1-tantheta=1+sectheta

Answer 1

Answer:

$\sf 1 \ + \ sec \theta cosec \theta \ = \ RHS$

Question :

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

Solution :

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

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

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

$\implies \sf \dfrac {tan^2 \theta}{tan \theta \ - \ 1} \ - \ \dfrac {1}{tan \theta (\theta \ - \ 1)}$

$\implies \sf \dfrac {tan^3 \theta \ - \ 1}{tan \theta (tan \theta \ - \ 1)}$

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

$\implies \sf \dfrac {tan^2 \theta + tan \theta \ + \ 1}{tan \theta}$

$\implies \sf \dfrac {tan^2 \theta}{tan \theta} \ + \ \dfrac {tan \theta}{tan \theta} \ + \ \dfrac {1}{tan \theta}$

$\implies \sf tan \theta \ + \ 1 \ + \ cot \theta$

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

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

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

$\implies \bf 1 \ + \ sec \theta \ cosec \theta \ = \ RHS$

Answer 2

$\pink{\sf{\star\;Answer}}$

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➠ (Secθ -Cosecθ)(1 + Tanθ + Cotθ) = TanθSecθ - CotθCosecθ

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$\orange{\sf{\star\;Step-\;by-\;step\; explanation:}}$

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➠ (Secθ -Cosecθ)(1 + Tanθ + Cotθ) = Tanθ Secθ - CotθCosecθ

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$\red{\sf{\star\;LHS}}$

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➠ (Secθ -Cosecθ)(1 + Tanθ + Cotθ)

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➠ (1/Cosθ - 1/Sinθ)(1  + Sinθ/Cosθ + Cosθ/Sinθ)

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➠ {(Sinθ - Cosθ)/(CosθSinθ) }{CosθSinθ  + Sin²θ + Cos²θ)CosθSinθ) }

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➠ (1 + CosθSinθ)(Sinθ - Cosθ))(Cos²θSin²θ)

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➠ (Sinθ - Cosθ  + Sin²θCosθ -Cos²θSinθ)/(Cos²θSin²θ)

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➠ (Sinθ(1 - Cos²θ)  -Cosθ(1 -Sin²θ)/(Cos²θSin²θ)

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➠ (SinθSin²θ - CosθCos²θ)(Cos²θSin²θ)

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➠ Sinθ/Cos²θ  - Cosθ/Sin²θ

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➠ Tanθ/Cosθ - Cotθ/Sinθ

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➠ TanθSecθ - CotθCosecθ

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➠ $\blue{\sf{\star\;RHS}}$

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$\purple{\sf{\star\;Proved!}}$

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(Secθ -Cosecθ)(1 + Tanθ + Cotθ) = TanθSecθ - CotθCosecθ${\boxed{\green{\checkmark{}}}}$

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