Question

Sin theta(1+tan theta) +cos theta (1+cot theta) =(sec theta+ cosec theta)

Answer 1

Answer:

We need to prove: $sin\,\theta(1+tan\,\theta)+cos\,\theta(1+cot\,\theta)=sec\,\theta+cosec\,\theta$

$LHS=sin\,\theta(1+tan\,\theta)+cos\,\theta(1+cot\,\theta)$

using, $tan\,\theta=\frac{sin\,\theta}{cos\,\theta}\:\:and\:\:cot\,\theta=\frac{cos\,\theta}{sin\,\theta}$

$=sin\,\theta(1+\frac{sin\,\theta}{cos\,\theta})+cos\,\theta(1+\frac{cos\,\theta}{sin\,\theta})$

$=sin\,\theta(\frac{cos\,\theta+sin\,\theta}{cos\,\theta})+cos\,\theta(\frac{sin\,\theta+cos\,\theta}{sin\,\theta})$

$=(cos\,\theta+sin\,\theta)(\frac{sin\,\theta}{cos\,\theta})+\frac{cos\,\theta}{sin\,\theta})$

$=(cos\,\theta+sin\,\theta)(\frac{sin^2\,\theta+cos^2\,\theta}{cos\,\theta\:sin\,\theta})$

We know that, $sin^2\,\theta+cos^2\,\theta=1$

$=(cos\,\theta+sin\,\theta)(\frac{1}{cos\,\theta\:sin\,\theta})$

$=\frac{cos\,\theta+sin\,\theta}{cos\,\theta\:sin\,\theta}$

$=\frac{cos\,\theta}{cos\,\theta\:sin\,\theta}+\frac{sin\,\theta}{cos\,\theta\:sin\,\theta}$

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

we know that, $\frac{1}{sin\,\theta}=cosec\,\theta\:\:and\:\:\frac{1}{cos\,\theta}=sec\,\theta$

$=cosec\,\theta+sec\,\theta$

= RHS

Hence Proved.