[SOLVED] Trig Identity proofs

raveen4706 · January 21st 2007, 05:17 AM

Could someone please prove secx - 1/ 1 - cos x = sec x and 1/ tanx + cotx = sinxcosx?

x = angle
Thanks in advance

**Soroban** · January 21st 2007, 06:39 AM

Hello, raveen4706!

You really must learn to use parentheses . . .

$\frac{\sec x - 1}{1 - \cos x}\; =\;\sec x$

The left side is: . $\frac{\frac{1}{\cos x} - 1}{1 - \cos x}$

Multiply top and bottom by $\cos x\!:\;\;\frac{\cos x}{\cos x}\cdot\frac{\frac{1}{\cos x} - 1}{1 - \cos x} \;=\;\frac{1 - \cos x}{\cos x(1 - \cos x)}$

Reduce: . $\frac{1}{\cos x} \;=\;\sec x$

$\frac{1}{\tan x + \cot x} \:= \:\sin x\cos x$

The left side is: . $\frac{1}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}$

Multiply top and bottom by $\sin x\cos x\!:\;\;\frac{\sin x\cos x}{\sin x\cos x}\cdot\frac{1}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}$

And we get: . $\frac{\sin x\cos x}{\sin^2\!x + \cos^2\!x} \;=\;\frac{\sin x\cos x}{1} \;=\;\sin x\cos x$

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