Question

# If trigonometric ratios $\cot \left( \theta -\alpha \right),3\cot \theta ,\cot \left( \theta +\alpha \right)$ are in AP and $\theta$ is not an integral multiple of $\dfrac{\pi }{2}$, then find the value of $\dfrac{4{{\sin }^{2}}\theta }{3{{\sin }^{2}}\alpha }$.

Hint: We will use various trigonometric identities to solve this question some of them are as states below,
$\cot \theta =\dfrac{\cos \theta }{\sin \theta },\sin 2\theta =2\sin \theta \cos \theta ,\sin \left( A+B \right)=\sin A.\cos B+\cos A.\sin B$ and $2\sin A.\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$. Also we will use the fact that if 3 numbers a, b & c are in AP then, $2b = a + c.$

Complete step-by-step solution:
Given that $\cot \left( \theta -\alpha \right),3\cot \theta ,\cot \left( \theta +\alpha \right)$ are in AP.
If three numbers are in AP, then; suppose a, b & c are the three numbers in AP then,
$b=\dfrac{a+c}{2}$ or $2b=a+c$ ------ (1)
Here, Let $a=\cot \left( \theta -\alpha \right),b=3\cot \theta ,c=\cot \left( \theta +\alpha \right)$.
Substituting these values in equation (1), as they are in AP we get,
$2\left[ 3\cot \theta \right]=\cot \left( \theta -\alpha \right)+\cot \left( \theta +\alpha \right)$
Now because we have, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$.
Converting $\cot \theta$ in terms of $\cos \theta$ & $\sin \theta$ in above equation we get,
$6\dfrac{\cos \theta }{\sin \theta }=\dfrac{\cos \left( \theta -\alpha \right)}{\sin \left( \theta -\alpha \right)}+\dfrac{\cos \left( \theta +\alpha \right)}{\sin \left( \theta +\alpha \right)}$
Now taking LCM of denominator we get,
$6\dfrac{\cos \theta }{\sin \theta }=\dfrac{\cos \left( \theta -\alpha \right)\sin \left( \theta +\alpha \right)+\cos \left( \theta +\alpha \right)\sin \left( \theta -\alpha \right)}{\sin \left( \theta -\alpha \right)\sin \left( \theta +\alpha \right)}$
Now we have a trigonometric identity as,
$\sin A.\cos B+\cos A.\sin B=\sin \left( A+B \right)$
Let, $A=\left( \theta -\alpha \right),B=\left( \theta +\alpha \right)$
Using this trigonometric identity in above equation we get,
\begin{align} & 6\dfrac{\cos \theta }{\sin \theta }=\dfrac{\sin \left( \theta +\alpha +\theta -\alpha \right)}{\sin \left( \theta -\alpha \right)\sin \left( \theta +\alpha \right)} \\ & \Rightarrow 6\dfrac{\cos \theta }{\sin \theta }=\dfrac{\sin \left( 2\theta \right)}{\sin \left( \theta +\alpha \right)\sin \left( \theta -\alpha \right)} \\ \end{align}
Using trigonometric identity, $\sin 2\theta =2\sin \theta \cos \theta$ in above we get,
$\Rightarrow 6\dfrac{\cos \theta }{\sin \theta }=\dfrac{2\sin \theta \cos \theta }{\sin \left( \theta +\alpha \right)\sin \left( \theta -\alpha \right)}$
Cross multiplying both we have,
$6\cos \theta \left[ \sin \left( \theta +\alpha \right)\sin \left( \theta -\alpha \right) \right]=2{{\sin }^{2}}\theta \cos \theta$
Cancelling $\cos \theta$ and 2 from both sides we get,
$3\left[ \sin \left( \theta +\alpha \right)\sin \left( \theta -\alpha \right) \right]={{\sin }^{2}}\theta$
Using trigonometric identity stated as,
$2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$ in above by taking, $A=\theta +\alpha$ and $B=\theta -\alpha$ we get,
\begin{align} & \Rightarrow 3\left[ \cos \left( \theta +\alpha -\theta +\alpha \right)-\cos \left( \theta -\alpha +\theta +\alpha \right) \right]=2{{\sin }^{2}}\theta \\ & \Rightarrow 3\left( \cos 2\alpha -\cos 2\theta \right)=2{{\sin }^{2}}\theta \\ \end{align}
Now we will use trigonometric identity as stated,
$\cos 2A=1-2{{\sin }^{2}}A$
Using this above we get,
\begin{align} & \Rightarrow 3\left[ 1-2{{\sin }^{2}}\alpha +2{{\sin }^{2}}\theta -1 \right]=2{{\sin }^{2}}\theta \\ & \Rightarrow -6{{\sin }^{2}}\alpha +6{{\sin }^{2}}\theta =2{{\sin }^{2}}\theta \\ \end{align}
\begin{align} & \Rightarrow -6{{\sin }^{2}}\alpha =2{{\sin }^{2}}\theta -6{{\sin }^{2}}\theta \\ & \Rightarrow -6{{\sin }^{2}}\alpha =-4{{\sin }^{2}}\theta \\ \end{align}
Dividing by $-{{\sin }^{2}}\alpha$ both sides we get,
$\Rightarrow +6=+4\dfrac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\alpha }$
Dividing by 3 both sides we get,
$\Rightarrow \dfrac{4{{\sin }^{2}}\theta }{3{{\sin }^{2}}\alpha }=2$
So, answer is, $\dfrac{4{{\sin }^{2}}\theta }{3{{\sin }^{2}}\alpha }=2$.

Note: Always remember that it is given that $\theta$ is not an integral multiple of $\dfrac{\pi }{2}$ therefore we can never use the formula of $\sin \left( \theta +\dfrac{\pi }{2} \right)$ or $\sin \left( \alpha +\dfrac{\pi }{2} \right)$ throughout the solution. Hence we have done all calculation without using these formulas.
$\Rightarrow 3\left[ \sin \left( \theta +\alpha \right)\sin \left( \theta -\alpha \right) \right]={{\sin }^{2}}\theta$