Question

complete the following activity to prove that ​

Answer 1

may hopethis will help you

Answer 2

proved

Step-by-step explanation:

given : $\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}= \sec\theta - \tan\theta$

here ,

LHS

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

$\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}= \sqrt{\frac{1-\sin\theta \times (1-\sin\theta)}{1+\sin\theta\times (1-\sin\theta)}}\\\\\sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2 \theta}}= \frac{1-\sin\theta}{\sqrt{1-\sin^2 \theta}}\\\\\frac{1-\sin\theta}{\sqrt{cos^2 \theta}}= \frac{1-\sin\theta}{cos \theta}...(1)$

RHS : $\sec\theta - \tan\theta$

$\frac{1}{\cos\theta}- \frac{\sin\theta }{\cos\theta}$

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

solved ..

#Learn more

https://brainly.in/question/4680129