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 $\dfrac{4{{\sin }^{2}}\theta }{3{{\sin }^{2}}\alpha }$.

Answer 1

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,

& 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{aligned}$$ Using trigonometric identity, $$\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{aligned} & \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{aligned}$$ Now we will use trigonometric identity as stated, $$\cos 2A=1-2{{\sin }^{2}}A$$ Using this above we get, $$\begin{aligned} & \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{aligned}$$ $$\begin{aligned} & \Rightarrow -6{{\sin }^{2}}\alpha =2{{\sin }^{2}}\theta -6{{\sin }^{2}}\theta \\\ & \Rightarrow -6{{\sin }^{2}}\alpha =-4{{\sin }^{2}}\theta \\\ \end{aligned}$$ Dividing by $$-{{\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 $$