Question

1 - sin theta upon 1 + sin theta is equal to sec theta minus 10 theta whole square

Answer 1

i hope it is help full for you

Answer 2

Answer and Explanation:

To prove :  $\frac{1-\sin \theta}{1 +\sin \theta}=(\sec \theta - \tan \theta)^{2}$

Proof :

Taking LHS,

$=\frac{1-\sin \theta}{1 +\sin \theta}$

$=\frac{1-\sin \theta}{1 +\sin \theta}\times \frac{1-\sin \theta}{1-\sin \theta}$

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

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

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

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

$=(\sec \theta-\tan\theta)^{2}$

=RHS

Hence proved.