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

Prove the trigonometric identity above.

Note by ☃☆Jn Chu♧♢
1 year, 11 months ago

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\(\large \frac{1- \cos \theta}{1 + \cos \theta}= \frac{1- (1-2 \sin^2 \frac{\theta}{2})}{1+(2 \cos^2 \frac{\theta}{2} -1)}= \frac{2 \sin ^2 \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}= \tan^2 \frac{\theta}{2} \) Ravi Dwivedi · 1 year, 11 months ago

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Hint: \( \cos\theta = 1 - 2\sin^2\frac\theta2 = 2\cos^2\frac\theta2-1\). Pi Han Goh · 1 year, 11 months ago

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