# Proving $\left(\tan^2x +\frac{1}{\tan x}\right) \left(\sin x+\cos x\right)=\left(\frac{1}{\cos x}\right)+\left(\frac{1}{sin x}\right)$

$$\left(\tan^2x +\frac{1}{\tan x}\right) \left(\sin x+\cos x\right)=\left(\frac{1}{\cos x}\right)+\left(\frac{1}{sin x}\right)$$

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

• You could try expressing everything in terms of sines and cosines, clear fractions, and see what you get. – Gerry Myerson Dec 4 '13 at 2:14
• I don't think this is true. See Wolfram – Calvin Lin Dec 4 '13 at 2:16
• It's false. Plug in $x=\pi/4$. – Doc Dec 4 '13 at 2:23
• Well, did you try to follow Gerry's suggestion above? – Doc Dec 4 '13 at 2:46
• Bet you anything you want that your $\tan^2{x}$ term should be $\tan{x}$. – Doc Dec 4 '13 at 2:51

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 $x$.