Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$

Question

Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$

Also it is a question of S.L. Loney's Plane Trignonometry

What I've tried by now:

\begin{align} & =\frac{1+\sin\theta-\sin(90-\theta)}{1+\cos(90-\theta)+\cos\theta} \\[10pt] & =\frac{1+2\cos45^\circ \sin(\theta-45^\circ)}{1+2\cos45^\circ \cos(45-\theta)} \end{align}

Cause I do know
\begin{align} & \sin c + \sin d = 2\sin\left(\frac{c+d}{2}\right)\cos\left(\frac{c-d}{2}\right) \\[10pt] \text{and } & \cos c + \cos d = 2\cos\left(\frac{c+d}{2}\right)\sin\left(\frac{c-d}{2}\right) \end{align}

I can't think of what to do next..

Answer 1

Let $\theta = 2 \phi$, then the thing to be proven is:

Prove that $$\frac{1 + \sin(2\phi) - \cos(2\phi)}{1 + \sin(2\phi) + \cos(2\phi)} = \tan(\phi)$$

Then use:

$$\sin(2\phi) = 2 \sin \phi \cos \phi$$ $$\cos(2\phi) = \cos^2 \phi - \sin^2 \phi$$

and:

$$\sin^2 \phi + \cos^2 \phi = 1$$

Answer 2

hint just use

$$1+\cos (X)=2\cos^2 (\frac {X}{2}) $$

$$1-\cos (X)=2\sin^2 (\frac {X}{2}) $$

Answer 3

HINT: use the tan-half angle substution $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}$$ $$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$ with $t=\tan(x/2)$

Answer 4

Write $t=\tan\frac{\theta}{2}$ so $\sin\theta=\frac{2t}{1+t^2},\,\cos\theta=\frac{1-t^2}{1+t^2}$. Hence $$\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}=\frac{1+t^2+2t-1+t^2}{1+t^2+2t+1-t^2}=\frac{2t+2t^2}{2+2t}=t.$$

Answer 5

$$ \begin{aligned} & \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\\=& \frac{(1-\cos \theta)+\sin \theta}{(1+\cos \theta)+\sin \theta}\\ =& \frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)+2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)+2 \sin \left(\frac{\theta}{2}\right) \operatorname{cor}\left(\frac{\theta}{2}\right)} \\ =& \frac{\sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)} \cdot \frac{\sin \left(\frac{\theta}{2}\right)+\cos \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)+\sin \left(\frac{\theta}{2}\right)} \\ =& \tan \left(\frac{\theta}{2}\right) \end{aligned} $$