Question

Write $\frac{sin \theta}{1+cos \theta}$ in the terms of $tan \frac{\theta}{2}$

Answer 1

Proof :

Now,

$\large{\frac{sin\theta}{1+cos\theta}}$

$\large{=\frac{2sin\frac{\theta}{2}cos\frac{\theta}{2}}{2 {cos}^{2}\frac{\theta}{2}}\{\because{cos}^{2}x-{sin}^{2}x=cos2x\}}$

$\large{=\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}}$

$\large{=tan\frac{\theta}{2}}$

$\implies\boxed{\bold{\frac{sin\theta}{1+cos\theta}=tan\frac{\theta}{2}}}$

Hence, proved.

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