Question

Prove that : cot^2 A(sec theta - 1)/1 + sin theta = sec^2 theta ( 1 - sin theta/1 + sec theta )​

Answer 1

hope it's helpful to you. . .

Answer 2

Answer:

To Prove: $\frac{cot^2\,\theta(sec\,\theta-1)}{(1+sin\,\theta)}=sec^2\,\theta(\frac{1-sin\,\theta}{1+sec\,\theta})$

Consider,

LHS

$=\frac{cot^2\,\theta(sec\,\theta-1)}{(1+sin\,\theta)}$

$=\frac{cot^2\,\theta(sec\,\theta-1\times\frac{sec\,\theta+1}{sec\,\theta+1})}{(1+sin\,\theta\times\frac{1-sin\,\theta}{1-sin\,\theta})}$

$=\frac{cot^2\,\theta(\frac{sec^2\,\theta+sec\,\theta-sec\,\theta-1}{sec\,\theta+1})}{\frac{1-sin\,\theta+sin\,\theta-sin^2\,\theta}{1-sin\,\theta}}$$=\frac{cot^2\,\theta(\frac{sec^2\,\theta-1}{sec\,\theta+1})}{\frac{1-sin^2\,\theta}{1-sin\,\theta}}$

$=\frac{cot^2\,\theta(\frac{tan^2\,\theta}{sec\,\theta+1})}{\frac{cos^2\,\theta}{1-sin\,\theta}}$

$=(\frac{cot^2\,\theta\times tan^2\,\theta}{sec\,\theta+1})(\frac{1-sin\,\theta}{cos^2\,\theta})$

$=\frac{1(1-sin\,\theta)}{(sec\,\theta+1)cos^2\,\theta}$

$=\frac{sec^2\,\theta(1-sin\,\theta)}{(sec\,\theta+1)}$

$=RHS$

Hence Proved.