Challenging Trig Identity Questions

**tmac20** · December 3rd 2011, 11:00 AM

Need some help with

$(9)\;\;1 - \frac{\sin^2\!x\tan x}{\tan x+1} - \frac{\cos^2\!x}{\tan x+1} \;=\;\sin x\cos x$

Thanks.

Moderator edit: Scanned images deleted, single question typed out.

**Soroban** · December 3rd 2011, 12:07 PM

Hello, tmac20!

Can you show us any of your work?

$(9)\;\;1 - \frac{\sin^2\!x\tan x}{\tan x+1} - \frac{\cos^2\!x}{\tan x+1} \;=\;\sin x\cos x$

We have: . $1 - \dfrac{\dfrac{\sin^2\!x}{1}\,\dfrac{\sin x}{\cos x}}{\dfrac{\sin x}{\cos x} + 1} - \dfrac{\cos^2\!x}{\dfrac{\sin x}{\cos x} + 1}$

Multiply both fractions by $\frac{\cos x}{\cos x}\!:\;\;1 - \frac{\sin^3\!x}{\sin x + \cos x} - \frac{\cos^3\!x}{\sin x+\cos x}$

. . $=\;\;1 - \frac{\overbrace{\sin^3\!x + \cos^3\!x}^{\text{sum of cubes}}}{\sin x+\cos x} \;\;=\;\;1 - \frac{(\sin x+\cos x)(\sin^2\!x - \sin x\cos x + \cos^2x)}{\sin x+\cos x}$

. . $=\;\;1 - (\underbrace{\sin^2\!x+\cos^2\!x}_{\text{This is 1}} - \sinx\cos x) \;\;=\;\; 1 - (1 - \sin x\cos x)$

. . $=\;\;1 - 1 + \sin x\cos x \;\;=\;\;\sin x\cos x$

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