Chapter 7.1, Problem 72E

To determine

To verify the identity:

  $\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v\sin v}$

Explanation of Solution

Given:

An identity is given as

  $\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v\sin v}$

Concept Used:

The trigonometric identities:

  $\begin{array}{l}\tan x=\frac{\sin x}{\cos x}\\ 1-{\cos }^{2}x={\sin }^{2}x\end{array}$

Algebraic identity

  $\begin{array}{l}{\left(a-b\right)}^2=a^2-2ab+b^2\\ a^2-b^2=\left(a-b\right)\left(a+b\right)\end{array}$

Calculation:

In order to verify the identity:

  $\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v\sin v}$

Take left hand of above identity and simplify it using the trigonometric identities.

As,

  $\begin{array}{c}\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\frac{\sin v}{\cos v}\times \sin v}{\frac{\sin v}{\cos v}+\sin v}\\ =\frac{\frac{{\sin }^{2}v}{\cos v}\times \cos v}{\sin v+\sin v\cos v}\\ =\frac{{\sin }^{2}v}{\sin v+\sin v\cos v}\\ =\frac{1-{\cos }^{2}v}{\sin v\left(1+\cos v\right)}\\ =\frac{\left(1+\cos v\right)\left(1-\cos v\right)}{\sin v\left(1+\cos v\right)}\\ =\frac{1-\cos v}{\sin v}\\ =\frac{1-\cos v}{\sin v}\times \frac{\sin v}{\sin v}\\ =\frac{\sin v-\sin v\cos v}{{\sin }^{2}v}\\ =\frac{\frac{\sin v-\sin v\cos v}{\cos v}}{\frac{{\sin }^{2}v}{\cos v}}\\ =\frac{\frac{\sin v}{\cos v}-\frac{\sin v\cos v}{\cos v}}{\frac{\sin v}{\cos v}\times \sin v}\\ =\frac{\tan v-\sin v}{\tan v\sin v}\\ =\text{Right hand}\end{array}$

Thus, identity is verified, i.e.

  $\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v\sin v}$