Question

(1+sin a-cosa)/(1+sin a+cos a)= tan a/2​

Answer 1

$\sf\bf\large\dfrac{1 + sin \theta - cos\theta}{1 + sin\theta + cos\theta} = tan \dfrac{\theta}{2} \\\\\\L.H.S \longrightarrow\ \dfrac{1 + sin \theta - cos\theta}{1 + sin\theta + cos\theta}\\\\\\Re-writing \: the \: Equation:\\\\\\\dfrac{1 - cos \theta + sin\theta}{1 + cos\theta + sin\theta} \longrightarrow Equation 1$

We know the following Trigonometric Identities. By Transposing them, We get;

• cos 2Θ = 2 cos²Θ -1

⇒ 2 cos²Θ = cos 2Θ + 1

⇒ 1 + cos 2Θ = 2 cos²Θ

⇒ $\sf\bf\large 1 + cos\: \theta = 2 cos^2 \dfrac{\theta}{2}$

• cos 2Θ = 1 - 2 sin²Θ

⇒ 2 sin²Θ = 1 - cos 2Θ

⇒ 1 - cos 2Θ = 2 sin²Θ

$\Rightarrow \bf{1 - cos \: \theta} = 2 sin^2 \dfrac{\theta}{2}}$

• sin 2Θ = 2 sinΘ cosΘ

$\sf\bf\large sin \theta = 2 \: sin \dfrac{\theta}{2} cos \dfrac{\theta}{2}$

Substituting in Equation 1:

$\sf\bf\large \dfrac{(1 - cos \theta) + sin\theta}{(1 + cos\theta) + sin\theta}$

$\dfrac{2 sin^2 \dfrac{\theta}{2} + 2 sin \dfrac{\theta}{2} cos \dfrac{\theta}{2} }{2 cos^2 \dfrac{\theta}{2} + 2 sin \dfrac{\theta}{2} cos \dfrac{\theta}{2} }$

Taking Common Factor out:

$\dfrac{2 sin \dfrac{\theta}{2} (sin \dfrac{\theta}{2} + cos \dfrac{\theta}{2} ) }{2 cos \dfrac{\theta}{2} (sin \dfrac{\theta}{2} + cos \dfrac{\theta}{2} )}$

Striking out the common numerator and denominator, we are left with:

$\dfrac{sin \dfrac{\theta}{2} }{cos \dfrac{\theta}{2} } \\\\\\\sf\bf\large tan \dfrac{\theta}{2} = R.H.S$

Hence proved.