Question

Proving Identities

Verify the identity$\frac{\tan v-\cos v}{{\tan }^{2}v-{\cos }^{2}v}=\sin v\cos v$

Short answer

The expression$\frac{\tan v-\cos v}{{\tan }^{2}v-{\cos }^{2}v}=\sin v\cos v$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression$\frac{\tan v-\cos v}{{\tan }^{2}v-{\cos }^{2}v}=\sin v\cos v$

Step 2. Concept used.

Divide the supplied statement into LHS and RHS sections. Simplify each one separately, then compare the results. Given expression is an identity if both outcomes are equivalent.

Step 3. Calculation.

Now, simplify LHS of $\frac{\tan v-\cos v}{{\tan }^{2}v-{\cos }^{2}v}=\sin v\cos v$:

Convert all into sine & cosine,

$\begin{array}{l}\frac{\tan v-\cos v}{{\tan }^{2}v-{\cos }^{2}v}=\frac{\frac{\sin v}{\cos v}-\frac{\cos v}{\sin v}}{\frac{{\sin }^{2}v}{{\cos }^{2}v}-\frac{{\cos }^{2}v}{{\sin }^{2}v}}\\ =\frac{\frac{{\sin }^{2}v-{\cos }^{2}v}{\cos v\sin v}}{\frac{{\sin }^{4}v-{\cos }^{4}v}{{\cos }^{2}v{\sin }^{2}v}}\\ =\frac{\frac{{\sin }^{2}v-{\cos }^{2}v}{\cos v\sin v}}{\frac{\left({\sin }^{2}v-{\cos }^{2}v\right)\left({\sin }^{2}v+{\cos }^{2}v\right)}{{\cos }^{2}v{\sin }^{2}v}}\\ \because {\sin }^2\theta +{\cos }^2\theta =1\\ =\frac{\frac{{\sin }^{2}v-{\cos }^{2}v}{\cos v\sin v}}{\frac{1\times \left({\sin }^{2}v-{\cos }^{2}v\right)}{{\cos }^{2}v{\sin }^{2}v}}\\ =\sin v\cos v\end{array}$

RHS of the equation is$\sin v\cos v$

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