Question #a4b0f

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Question #a4b0f

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Answer 1 · 2017-10-08T08:24:32

proved $\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + tan θ + cot θ$ $tan theta/(1-cot theta) + cot theta/(1-tan theta) = 1+tan theta+cot theta$

Explanation:

We have to prove, $\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + tan θ + cot θ$ $tan theta/(1-cot theta) + cot theta/(1-tan theta) = 1+tan theta+ cot theta$

Let us take Left Hand Side (L.H.S.)
$\Rightarrow \frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ}$ $rArr tan theta/(1-cot theta) +cot theta/(1-tan theta)$

$\Rightarrow \frac{\frac{sin θ}{cos θ}}{1 - \frac{cos θ}{sin θ}} + \frac{\frac{cos θ}{sin θ}}{1 - \frac{sin θ}{cos θ}}$ $rArr [sin theta/cos theta]/[1-cos theta/sin theta] + [cos theta/sin theta]/[1-sin theta/cos theta]$

$\Rightarrow \frac{\frac{sin θ}{cos θ}}{\frac{sin θ - cos θ}{sin θ}} + \frac{\frac{cos θ}{sin θ}}{\frac{cos θ - sin θ}{cos θ}}$ $rArr [sin theta/cos theta]/[(sin theta-cos theta)/sin theta] + [cos theta/sin theta]/[(cos theta-sin theta)/cos theta]$

$\Rightarrow \frac{\frac{sin θ}{cos θ} . sin θ}{sin θ - cos θ} + \frac{\frac{cos θ}{sin θ} . cos θ}{- (sin θ - cos θ)}$ $rArr [sin theta/cos theta. sin theta]/(sin theta-cos theta) +[cos theta/sin theta. cos theta]/[-(sin theta-cos theta)]$

$\Rightarrow \frac{\frac{{sin}^{2} θ}{cos θ}}{sin θ - cos θ} - \frac{\frac{{cos}^{2} θ}{sin θ}}{sin θ - cos θ}$ $rArr [sin^2 theta/cos theta]/[sin theta-cos theta] - [cos^2 theta/sin theta]/[sin theta-cos theta]$

$\Rightarrow \frac{\frac{{sin}^{2} θ}{cos θ} - \frac{{cos}^{2} θ}{sin θ}}{sin θ - cos θ}$ $rArr [sin^2 theta/cos theta - cos^2 theta/sin theta]/(sin theta-cos theta)$

$\Rightarrow \frac{\frac{{sin}^{3} θ - {cos}^{3} θ}{sin θ cos θ}}{sin θ - cos θ}$ $rArr [(sin^3 theta - cos^3 theta)/(sin theta cos theta)]/(sin theta-cos theta)$

$\Rightarrow \frac{(sin θ - cos θ) ({sin}^{2} θ + sin θ . cos θ + {cos}^{2} θ)}{sin θ cos θ} . \frac{1}{sin θ - cos θ}$ $rArr [(sin theta-cos theta)(sin^2 theta+sin theta. cos theta+cos^2 theta)]/(sin theta cos theta). 1/(sin theta-cos theta)$

$\Rightarrow \frac{{sin}^{2} θ}{sin θ cos θ} + \frac{sin θ . cos θ}{sin θ . cos θ} + \frac{{cos}^{2} θ}{sin θ . cos θ}$ $rArr sin^2 theta/(sin theta cos theta)+(sin theta. cos theta)/(sin theta. cos theta)+cos^2 theta/(sin theta. cos theta)$

$\Rightarrow \frac{sin θ}{cos θ} + 1 + \frac{cos θ}{sin θ}$ $rArr sin theta/cos theta +1+cos theta/sin theta$

$\Rightarrow tan θ + 1 + cot θ$ $rArr tan theta+1+cot theta$ = L. H. S.

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