Question

Proving Identities

Verify the identity$\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v\sin v}$

Short answer

The expression$\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v\sin v}$ is an identity.

Step-by-step solution

Step 1. Given information

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

Step 2. Concept used

For such type of question, split the expression into two subparts as LHS and RHS. Simplify them separately. If the results of both are equal then it is an identity.

Step 3. Calculation

Now, simplify LHS of $\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v\sin v}$

$\begin{array}{l}\frac{\tan v\sin v}{\tan v+\sin v}=\frac{\frac{\sin v}{\cos v}\sin v}{\frac{\sin v}{\cos v}+\sin v}\\ \left(\because \tan v=\frac{\sin v}{\cos v}\right)\\ =\frac{\frac{{\sin }^{2}v}{\cos v}}{\frac{\sin v+\sin v\times \cos v}{\cos v}}\\ =\frac{{\sin }^{2}v}{\sin v(1+\cos v)}\\ =\frac{1-{\cos }^{2}v}{\sin v(1+\cos v)}\end{array}$

Multiply and divide by $1-\cos v$

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

RHS of the equation is$\frac{\tan v-\sin v}{\tan v\sin v}$

Both RHS and LHS are equal. Hence it is an identity