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

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.
\[\Rightarrow 3\left[ \sin \left( \theta +\alpha \right)\sin \left( \theta -\alpha \right) \right]={{\sin }^{2}}\theta \]