Question

show that secant theta minus tan theta whole square equals to 1 -sin theta /1 + sin theta

Answer 1

1-sin/1+sin *1-sin/1-sin
(1-sin)^2/1-sin^2
(1-sin)^2/cos^2
[(1/cos)-(sin/cos)]^2
(sectheta-tan theta)^2
=lhs

Answer 2

Answer:

The proof is explained step-wise below :

Step-by-step explanation:

$\text{To Prove : }\sec\theta-\tan\theta=\frac{1-\sin\theta}{1+\sin\theta}$

Taking R.H.S first :

$\implies\frac{1-\sin\theta}{1+\sin\theta}\\\\\text{Rationalizing the above fraction}\\\\\implies \frac{1-\sin\theta}{1+\sin\theta}\times \frac{1-\sin\theta}{1-\sin\theta}\\\\\implies \frac{(1-\sin\theta)^2}{1-\sin^2\theta}\\\\\implies \frac{(1-\sin\theta)^2}{\cos^2\theta}\\\\\implies(\frac{1-\sin\theta}{\cos\theta})^2\\\\\implies (\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta})^2\\\\\implies(\sec\theta-\tan\theta)^2$

= R.H.S.

Hence Proved