Chapter 7, Problem 51RE

To determine

To establish:the given identity.

Answer to Problem 51RE

The identity is established.

Explanation of Solution

Given:

The given identity is $\frac{\csc \theta }{1+\csc \theta }=\frac{1-\sin \theta }{{\cos }^2\theta }$ .

Calculation:

As the given identity is $\frac{\csc \theta }{1+\csc \theta }=\frac{1-\sin \theta }{{\cos }^2\theta }$ .

Now, let us check LHS:

  $\begin{array}{l}\Rightarrow LHS=\frac{\csc \theta }{1+\csc \theta }\\ =\frac{\frac{1}{\sin \theta }}{1+\frac{1}{\sin \theta }}\\ =\frac{\frac{1}{\sin \theta }\times \sin \theta }{\left(1+\frac{1}{\sin \theta }\right)\times \sin \theta }\\ =\frac{1}{\sin \theta +1}\end{array}$

  $\begin{array}{l}RHS=\frac{1-\sin \theta }{{\cos }^2\theta }\\ =\frac{1-\sin \theta }{{\cos }^2\theta }\\ \left\{\text{since, }{\cos }^2\theta =1-{\sin }^2\theta \right\}\\ =\frac{1-\sin \theta }{1-{\sin }^2\theta }\\ =\frac{1-\sin \theta }{\left(1-\sin \theta \right)\left(1+\sin \theta \right)}\\ =\frac{1}{\left(1+\sin \theta \right)}\\ \therefore LHS=RHS\end{array}$

Thus, $\frac{\csc \theta }{1+\csc \theta }=\frac{1-\sin \theta }{{\cos }^2\theta }$

Hence, the identity is established.