Question

Solution of the equation \[3\tan \left( \theta -{{15}^{\circ }} \right)=\tan \left( \theta +{{15}^{\circ }} \right)\] is
1. \[\theta =n\pi -\dfrac{\pi }{3}\]
2. \[\theta =n\pi +\dfrac{\pi }{3}\]
3. \[\theta =n\pi -\dfrac{\pi }{4}\]
4. \[\theta =\dfrac{n\pi }{2}+{{(-1)}^{n}}\dfrac{\pi }{4}\]

Answer 1

Hint: Here we have been given a trigonometric equation and we have to find the value of the angle in it. Firstly we will take the constant term one side and the trigonometric value on the other side. Then we will use componendo and dividendo for simplifying our terms. Finally by using the addition and subtraction formula of trigonometry we will simplify our value and get the desired answer.

Complete answer: We have to find the solution of,
\[3\tan \left( \theta -{{15}^{\circ }} \right)=\tan \left( \theta +{{15}^{\circ }} \right)\]
Taking the constant on one side and trigonometric value on another we get,
\[\Rightarrow \dfrac{\tan \left( \theta +{{15}^{\circ }} \right)}{\tan \left( \theta -{{15}^{\circ }} \right)}=\dfrac{3}{1}\]
The componendo and dividendo formula is given as,
For any $\dfrac{a}{b}=\dfrac{c}{d}$ we can write it as $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$
Using above formula we get,
\[\Rightarrow \dfrac{\tan \left( \theta +{{15}^{\circ }} \right)+\tan \left( \theta -{{15}^{\circ }} \right)}{\tan \left( \theta +{{15}^{\circ }} \right)-\tan \left( \theta -{{15}^{\circ }} \right)}=\dfrac{3+1}{3-1}\]
\[\Rightarrow \dfrac{\tan \left( \theta +{{15}^{\circ }} \right)+\tan \left( \theta -{{15}^{\circ }} \right)}{\tan \left( \theta +{{15}^{\circ }} \right)-\tan \left( \theta -{{15}^{\circ }} \right)}=\dfrac{4}{2}\]
So we get,
\[\dfrac{\tan \left( \theta +{{15}^{\circ }} \right)+\tan \left( \theta -{{15}^{\circ }} \right)}{\tan \left( \theta +{{15}^{\circ }} \right)-\tan \left( \theta -{{15}^{\circ }} \right)}=2\]…..$\left( 1 \right)$
Now as we know tangent can be written in form of sine and cosine as follows,
$\tan A=\dfrac{\sin A}{\cos A}$
Using it in equation (1) we get,
\[\Rightarrow \dfrac{\dfrac{\sin \left( \theta +{{15}^{\circ }} \right)}{\cos \left( \theta +{{15}^{\circ }} \right)}+\dfrac{\sin \left( \theta -{{15}^{\circ }} \right)}{\cos \left( \theta -{{15}^{\circ }} \right)}}{\dfrac{\sin \left( \theta +{{15}^{\circ }} \right)}{\cos \left( \theta +{{15}^{\circ }} \right)}-\dfrac{\sin \left( \theta -{{15}^{\circ }} \right)}{\cos \left( \theta -{{15}^{\circ }} \right)}}=2\]
Taking L.C.M in numerator and denominator we get,
\[\Rightarrow \dfrac{\dfrac{\sin \left( \theta +{{15}^{\circ }} \right)\cos \left( \theta -{{15}^{\circ }} \right)+\sin \left( \theta -{{15}^{\circ }} \right)\cos \left( \theta +{{15}^{\circ }} \right)}{\cos \left( \theta +{{15}^{\circ }} \right)\cos \left( \theta -{{15}^{\circ }} \right)}}{\dfrac{\sin \left( \theta +{{15}^{\circ }} \right)\cos \left( \theta -{{15}^{\circ }} \right)-\sin \left( \theta -{{15}^{\circ }} \right)\cos \left( \theta +{{15}^{\circ }} \right)}{\cos \left( \theta +{{15}^{\circ }} \right)\cos \left( \theta -{{15}^{\circ }} \right)}}=2\]
Cancelling the common value of numerator and denominator we get,
\[\Rightarrow \dfrac{\sin \left( \theta +{{15}^{\circ }} \right)\cos \left( \theta -{{15}^{\circ }} \right)+\sin \left( \theta -{{15}^{\circ }} \right)\cos \left( \theta +{{15}^{\circ }} \right)}{\sin \left( \theta +{{15}^{\circ }} \right)\cos \left( \theta -{{15}^{\circ }} \right)-\sin \left( \theta -{{15}^{\circ }} \right)\cos \left( \theta +{{15}^{\circ }} \right)}=2\]…..$\left( 2 \right)$
Next we know the Addition and subtraction formula for sine is as follows,
\[\sin (A+B)=\sin A\cos B+\cos A\sin B\]
\[\sin (A-B)=\sin A\cos B-\cos A\sin B\]
Using the above formula in equation (2) where $A=\theta $ and $B={{15}^{\circ }}$ we get,
\[\Rightarrow \dfrac{\sin (\theta +{{15}^{\circ }}+\theta -{{15}^{\circ }})}{\sin (\theta +{{15}^{\circ }}-\theta +{{15}^{\circ }})}=2\]
$\Rightarrow \dfrac{\sin (2\theta )}{\sin ({{30}^{\circ }})}=2$
We know that $\sin {{30}^{\circ }}=\dfrac{1}{2}$ substitute above,
$\Rightarrow \dfrac{\sin (2\theta )}{\dfrac{1}{2}}=2$
$\Rightarrow 2\sin \left( 2\theta \right)=2$
Simplifying we get,
$\Rightarrow \sin \left( 2\theta \right)=1$
We know $\sin \dfrac{\pi }{2}=1$ so we get,
$\Rightarrow \sin \left( 2\theta \right)=\sin \dfrac{\pi }{2}$
We know that general solution of above term is calculated as
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $
Where $\alpha =$ Angle on the right side and $\theta =$ angle on the left side so we get,
$2\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{2}$
$\Rightarrow \theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$
Hence the correct answer is (4).

Note:
This type of question required very much attention as the calculation is big there is a chance of error in it. Always write each step so that we don’t get confused and don’t make any silly mistake. First step when we have cotangent or tangent function is to change them into sine and cosine function as they are easier to find. The componendo and dividendo method is used to form term whose formula is already known to us.