Another trig identity

**Jkeeb** · February 14th 2011, 04:18 PM

Very Helpful thank you very much.

There is one more I am struggling with however.

sec^2x cscx = cscx +secx tanx

Any ideas?

**Soroban** · February 14th 2011, 04:54 PM

Hello, Jkeeb!

We can prove it in both directions.

$\sec^2\!x\csc x \:=\: \csc x + \sec x\tan x$

The left side is: . $\sec^2\!x\csc x$

. . . . . . . . . . $=\;(1 + \tan^2\!x)\csc x$

. . . . . . . . . . $=\;\csc x + \tan^2\!x\csc x$

. . . . . . . . . . $\displaystyle =\;\csc x + \frac{\sin^2\!x}{\cos^2\!x}\!\cdot\!\frac{1}{\sin x}$

. . . . . . . . . . $\displaystyle =\;\csc x + \frac{1}{\cos x}\!\cdot\!\frac{\sin x}{\cos x}$

. . . . . . . . . . $=\;\csc x + \sec x\tan x$

The right side is: . $\csc x + \sec x\tan x$

. . . . . . . . . . . $\displaystyle =\;\frac{1}{\sin x} + \frac{1}{\cos x}\!\cdot\!\frac{\sin x}{\cos x}$

. . . . . . . . . . . $\displaystyle =\;\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}}}{\cos^2\!x\sin x}$

. . . . . . . . . . . $\displaystyle =\;\frac{1}{\cos^2\!x\sin x}$

. . . . . . . . . . . $\displaystyle =\;\frac{1}{\cos^2\!x}\!\cdot\!\frac{1}{\sin x}$

. . . . . . . . . . . $=\; \sec^2\!x\csc x$

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