Question

If angle $\theta$ is divided into two parts such that the tangents of one part is $\lambda$ times the tangent of other, and $\varphi$ is their difference, then show that sin $\theta =\frac{\lambda +1}{\lambda -1}sin\varphi$.

Answer 1

Lets divide the angle $\theta$ into the two angles ${\theta }_1$ and ${\theta }_2$
Hence, $\theta =\theta +1+{\theta }_2$ and lets define,
$\varphi ={\theta }_2-{\theta }_1$
$\Rightarrow tan{\theta }_2=\lambda tan\theta$
$\Rightarrow sin{\theta }_2=\lambda sin{\theta }_{1}cos{\theta }_2$
Now, lets then state :
$sin\theta =ksin\theta$
Using the definations, we may write
$sin({\theta }_2-{\theta }_1)=ksin(\theta +1+{\theta }_2)$
Using the angle sum and difference identities for sine, we obtain
$sin\left({\theta }_2\right)cos\left({\theta }_1\right)-cos\left({\theta }_2\right)sin\left({\theta }_1\right)=ksin\left({\theta }_2\right)cos\left({\theta }_1\right)+kcos$ $\left({\theta }_2\right)sin\left({\theta }_1\right)(l-k)sin{\theta }_{2}cos{\theta }_1=(l+k)sin{\theta }_{1}cos{\theta }_2$ using sin ${\theta }_1=\lambda sin{\theta }_{1}cos{\theta }_2$
we get
$\lambda (l-k)=l+k$
$\lambda -\lambda k=l+k$
$\lambda -1=k(\lambda +1)$
$k=\frac{\lambda -1}{\lambda +1}$
$\frac{sin\varphi }{sin\theta }=\frac{\lambda -1}{\lambda +1}$
$sin\theta =\frac{\lambda -1}{\lambda +1}sin\varphi$
Hence proved.