Question

Question 5 (iii)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(iii) $\frac{tan\theta }{(1-cot\theta )}+\frac{cot\theta }{(1-tan\theta )}=1+sec\theta cosec\theta$
[Hint : Write the expression in terms of $sin\theta$ and $cos\theta$]

Answer 1

(iii)$\frac{tan\theta }{(1-cot\theta )}+\frac{cot\theta }{(1-tan\theta )}=1+sec\theta cosec\theta$
$L.H.S.=\frac{tan\theta }{(1-cot\theta )}+\frac{cot\theta }{(1-tan\theta )}=\left[\frac{\left(\frac{sin\theta }{cos\theta }\right)}{1-\left(\frac{cos\theta }{sin\theta }\right)}\right]+\left[\frac{\left(\frac{cos\theta }{sin\theta }\right)}{1-\left(\frac{sin\theta }{cos\theta }\right)}\right]=\left[\frac{\left(\frac{sin\theta }{cos\theta }\right)}{\frac{(sin\theta -cos\theta )}{sin\theta }}\right]+\left[\frac{\frac{(cos\theta }{sin\theta })}{\frac{(cos\theta -sin\theta )}{cos\theta }}\right]$
$=\left[\frac{si{n}^2\theta }{cos\theta (sin\theta -cos\theta )}\right]+\left[\frac{co{s}^2\theta }{sin\theta (cos\theta -sin\theta )}\right]=\left[\frac{si{n}^2\theta }{cos\theta (sin\theta -cos\theta )}\right]-\left[\frac{co{s}^2\theta }{sin\theta (sin\theta -cos\theta )}\right]=\frac{1}{(sin\theta -cos\theta )}\left[\right(\frac{si{n}^2\theta }{cos\theta })-(\frac{co{s}^2\theta }{sin\theta }\left)\right]=\frac{1}{(sin\theta -cos\theta )}\times \left[\frac{(si{n}^3\theta -co{s}^3\theta )}{sin\theta cos\theta }\right]$
$=\frac{\left[\right(sin\theta -cos\theta \left)\right(si{n}^2\theta +co{s}^2\theta +sin\theta cos\theta \left)\right]}{\left[\right(sin\theta -cos\theta \left)sin\theta cos\theta \right]}=\frac{(1+sin\theta cos\theta )}{sin\theta cos\theta }=\frac{1}{sin\theta cos\theta }+1=1+sec\theta cosec\theta =R.H.S.$