Question

sin theta /1+cos theta =cosec theta - cot theta​

Answer 1

We are familiar with the identity given below.

$\large \csc^2\theta - \cot^2\theta=1$

And also, we can consider the following.

$\large \dfrac{1+\cos\theta}{1+\cos\theta}=1$

Since, both equal 1, we can say,

$\dfrac{1+\cos\theta}{1+\cos\theta}=\csc^2\theta -\cot^2\theta$

And we have to have,

$1.\ \ \sin\theta \cdot\csc\theta =1\\ \\ 2.\ \ \dfrac{\sin\theta}{\cos\theta}=\tan\theta\ \ \ \Longrightarrow\ \ \ \dfrac{\sin\theta}{\tan\theta}=\cos\theta\ \ \ \Longrightarrow\ \ \ \sin\theta\cdot\cot\theta=\cos\theta$

Now,

$\begin{aligned}&\frac{1+\cos\theta}{1+\cos\theta}=\csc^2\theta-\cot^2\theta\\ \\ \Longrightarrow\ \ &\frac{\sin\theta\cdot\csc\theta + \sin\theta\cdot\cot\theta}{1+\cos\theta}=(\csc\theta+\cot\theta)(\csc\theta-\cot\theta)\\ \\ \Longrightarrow\ \ &\frac{\sin\theta(\csc\theta+\cot\theta)}{1+\cos\theta}=(\csc\theta+\cot\theta)(\csc\theta-\cot\theta)\\ \\ \Longrightarrow\ \ &\large\boxed{\frac{\sin\theta}{1+\cos\theta}=\csc\theta-\cot\theta}\end{aligned}$

Hence Proved!!!