Question

Prove that
$\frac{tan\theta }{1-cot\theta }+\frac{cot\theta }{1-tan\theta }=1+tan\theta +cot\theta$

Answer 1

L.H.S = $\frac{tan\theta }{1-cot\theta }+\frac{cot\theta }{1-tan\theta }$

$=\frac{\frac{sin\theta }{cos\theta }}{1-\frac{cos\theta }{sin\theta }}+\frac{\frac{cos\theta }{sin\theta }}{1-\frac{sin\theta }{cos\theta }}$

$=\frac{\frac{sin\theta }{cos\theta }}{\frac{sin\theta -cos\theta }{sin\theta }}+\frac{\frac{cos\theta }{sin\theta }}{\frac{cos\theta -sin\theta }{cos\theta .}}$

$=\frac{si{n}^2\theta }{cos\theta (sin\theta -cos\theta )}+\frac{co{s}^2\theta }{sin(cos\theta -sin\theta )}$

$=\frac{si{n}^2\theta }{cos\theta (sin\theta -cos\theta )}-\frac{co{s}^2\theta }{sin\theta (sin\theta -cos\theta )}$

$=\frac{si{n}^3\theta -co{s}^3\theta }{sin\theta cos\theta (sin\theta -cos\theta )}$

$=\frac{(sin\theta -cos\theta )(si{n}^2\theta +sin\theta cos\theta +co{s}^2\theta )}{sin\theta cos\theta (sin\theta -cos\theta )}$

$-\frac{si{n}^2\theta +sin\theta cos\theta +co{s}^2\theta }{sin\theta cos\theta }$

$=\frac{si{n}^2\theta }{sin\theta cos\theta }+\frac{sin\theta cos\theta }{sin\theta cos\theta }+\frac{co{s}^2\theta }{sin\theta cos\theta }$

= $tan\theta +1++cot\theta$

$=1+tan\theta +cot\theta$

= R.H.S.