[Math] Proving $\left(\tan x +\frac{1}{\tan x}\right) \left(\sin x+\cos x\right)=\frac{1}{\cos x}+\frac{1}{\sin x}$

Question

$$
\left(\tan x +\frac{1}{\tan x}\right) \left(\sin x+\cos x\right)=\frac{1}{\cos x}+\frac{1}{\sin x}
$$

Can someone show me how to solve this identity and also explain the steps?

Answer 1

Multiplying out on the left hand side you get: $(tan + \frac{1}{tan})\cdot (sin+cos) \\ = (\frac{sin}{cos}+\frac{cos}{sin})\cdot (sin + cos) \\ = \frac{sin^2}{cos}+ cos + sin + \frac{cos^2}{sin} \\ = \frac{sin^2}{cos}+\frac{cos^2}{cos}+\frac{sin^2}{sin}+\frac{cos^2}{sin} \\ = \frac{1}{cos}+\frac{1}{sin}.$

Hence, $(tan + \frac{1}{tan})\cdot (sin+cos) = \frac{1}{cos}+\frac{1}{sin}$. Now just apply these operators to $x$.