Question

sin square theta tan theta + cos square theta cot theta + 2 sin theta cos theta equal tan theta + cot theta.
we have to prove it

Answer 1

Your answer is placed above

Answer 2

Answer:

To prove: $sin^2\,\theta\:tan\,\theta+cos^2\,\theta\:cot\,\theta+2sin\,\theta\:cos\,\theta=tan\,\theta+cot\,\theta$

Consider,

LHS

$=sin^2\,\theta\:tan\,\theta+cos^2\,\theta\:cot\,\theta+2sin\,\theta\:cos\,\theta$      

$=sin^2\,\theta\:tan\,\theta+cos^2\,\theta\:cot\,\theta+sin\,\theta\:cos\,\theta+sin\,\theta\:cos\,\theta$      

$=sin^2\,\theta\:tan\,\theta+sin\,\theta\:cos\,\theta+cos^2\,\theta\:cot\,\theta+sin\,\theta\:cos\,\theta$

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

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

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

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

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

$=tan\,\theta+cot\,\theta$

= RHS

Hence Proved