Question

establish the identity
\\(\frac{4\cot u - \tan u}{\cot u+\tan u}+1 = 5\cos^{2}u\\)
write the left side in terms of sine and cosine.
\\(\frac{}{}+1 (do not simplify)

Explanation:

Step1: Recall cot and tan definitions

Recall that $\cot u=\frac{\cos u}{\sin u}$ and $\tan u = \frac{\sin u}{\cos u}$. Substitute these into the left - hand side fraction.
The left - hand side fraction $\frac{4\cot u-\tan u}{\cot u+\tan u}$ becomes $\frac{4\frac{\cos u}{\sin u}-\frac{\sin u}{\cos u}}{\frac{\cos u}{\sin u}+\frac{\sin u}{\cos u}}$.
So the left - hand side of the identity $\frac{4\cot u - \tan u}{\cot u+\tan u}+1$ written in terms of sine and cosine is $\frac{4\frac{\cos u}{\sin u}-\frac{\sin u}{\cos u}}{\frac{\cos u}{\sin u}+\frac{\sin u}{\cos u}}+1$.

Answer:

$\frac{4\frac{\cos u}{\sin u}-\frac{\sin u}{\cos u}}{\frac{\cos u}{\sin u}+\frac{\sin u}{\cos u}}+1$