Question

1/1-sin theta +1/1+sin theta=? ​

Answer 1

$\sf \dfrac{1}{1-sin\theta}+\dfrac{1}{1+sin\theta}$

$=\sf (\dfrac{1}{1-sin\theta}\times \dfrac{1+sin\theta}{1+sin\theta}) + (\dfrac{1}{1+sin\theta}\times \dfrac{1-sin\theta}{1-sin\theta})$

$=\sf \dfrac{1+sin\theta}{1-sin^2\theta}+\dfrac{1-sin\theta}{1-sin^2\theta}$

$=\sf \dfrac{1+sin\theta}{cos^2\theta}+\dfrac{1-sin\theta}{cos^2\theta}$

$=\sf \dfrac{1+sin\theta+1-sin\theta}{cos^2\theta}$

$=\sf \dfrac{1+1}{cos^2\theta}$

$=\sf \dfrac{2}{cos^2\theta}$

$=\sf 2\times\dfrac{1}{cos^2\theta}$

$=\sf 2\:sec^2\theta$

Therefore,

$\boxed{\bf \dfrac{1}{1-sin\theta}+\dfrac{1}{1+sin\theta}=2\:sec^2\theta}$