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# Solve this anyone?

$\large \tan^2\frac \theta2= \frac{1-\cos\theta}{1+\cos\theta}$

Prove the trigonometric identity above.

Note by ☃☆Jn Chu♧♢
2 years, 4 months ago

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$$\large \frac{1- \cos \theta}{1 + \cos \theta}= \frac{1- (1-2 \sin^2 \frac{\theta}{2})}{1+(2 \cos^2 \frac{\theta}{2} -1)}= \frac{2 \sin ^2 \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}= \tan^2 \frac{\theta}{2}$$

- 2 years, 4 months ago

Hint: $$\cos\theta = 1 - 2\sin^2\frac\theta2 = 2\cos^2\frac\theta2-1$$.

- 2 years, 4 months ago