Question

Find the value of the expression $\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)$ is equal to
a) 3 tan 3A
b) tan 3A
c) cot 3A
d) sin 3A

Answer 1

Hint: Use trigonometric identities to simplify the relation. Convert given tan functions to sin and cos functions. Take any two tan functions to solve the given expression easily.

Trigonometric expression given here is;
$\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)$
Let, we represent the above relation by ‘M’, hence;
$M=\tan A+\tan \left( 60+A \right)-\tan \left( 60-A \right)$……………….. (1)
Let us solve last two terms by converting tan to sin and cos using relation;
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ …………………. (2)
Hence, equation (1), can be written as;
$M=\tan A+\dfrac{\sin \left( 60+A \right)}{\cos \left( 60+A \right)}-\dfrac{\sin \left( 60-A \right)}{\cos \left( 60-A \right)}$
Taking L.C.M in last two fractions, we get
$M=\tan A+\dfrac{\left( \sin \left( 60+A \right).\cos \left( 60-A \right)-\sin \left( 60-A \right).\cos \left( 60+A \right) \right)}{\cos \left( 60+A \right).\cos \left( 60-A \right)}$
Now, we know the trigonometric identities as;
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ or vice – versa can also be used.
And, we also know that
$\cos \left( A-B \right)\cos \left( A+B \right)={{\cos }^{2}}B-{{\sin }^{2}}A$
Using the above trigonometric identities in for simplifying the relation ‘M’ we get;
$\begin{aligned} & M=\tan A+\dfrac{\sin {\left( 60+A \right)-\left( 60-A \right)}}{{{\cos }^{2}}A-{{\sin }^{2}}{60}} \\\ & M=\tan A+\dfrac{\sin 2A}{{{\cos }^{2}}A-{{\sin }^{2}}60{}^\circ } \\\ \end{aligned}$
Now, we know the value of $sin\text{ }60{}^\circ$i.e. $\dfrac{\sqrt{3}}{2}$. Putting value of ${{\sin }^{2}}60$, in above equation, we get;
$\begin{aligned} & M=\tan A+\dfrac{\sin 2A}{{{\cos }^{2}}A-\dfrac{3}{4}} \\\ & or \\\ & M=\tan A+\dfrac{4\sin 2A}{4{{\cos }^{2}}A-3} \\\ \end{aligned}$
Now, we can use trigonometric identity $\sin 2A=2\sin A\cos A\text{ and }\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to simplify the above relation more, we get;
$M=\dfrac{\sin A}{\cos A}+\dfrac{8\sin A\cos A}{4{{\cos }^{2}}A-3}$
Now, taking ‘sinA’ common from both the terms and L.C.M as well, we get;
$\begin{aligned} & M=\sin A\left( \dfrac{1}{\cos A}+\dfrac{8\cos A}{4{{\cos }^{2}}A-3} \right) \\\ & or \\\ & M=\sin A\left( \dfrac{4{{\cos }^{2}}A-3+8{{\cos }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\\ & M=\sin A\left( \dfrac{12{{\cos }^{2}}A-3}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\\ \end{aligned}$
Now, using the relation ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\ or\ {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$ to replace ${{\cos }^{2}}A\ \text{ by}\ 1-{{\sin }^{2}}A$, we get;
$\begin{aligned} & M=\sin A\left( \dfrac{12\left( 1-{{\sin }^{2}}A \right)-3}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\\ & M=\sin A\left( \dfrac{9-12{{\sin }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\\ \end{aligned}$
Taking ‘3’ common from numerator, we get;
$\begin{aligned} & M=3\sin A\left( \dfrac{3-4{{\sin }^{2}}A}{\cos A\left( 4{{\cos }^{2}}A-3 \right)} \right) \\\ & or \\\ & M=3\sin A\left( \dfrac{3\sin A-4{{\sin }^{3}}A}{4{{\cos }^{3}}A-3\cos A} \right) \\\ \end{aligned}$
AS, we know trigonometric identities for converting $3\theta \ to\ \theta$ with sin and cos terms as follows;
$\begin{aligned} & \sin 3A=3\sin A-4{{\sin }^{3}}A \\\ & \cos 3A=4{{\cos }^{3}}A-3\cos A \\\ \end{aligned}$
Hence, above relation of M can be re-written as;
$M=3\dfrac{\sin 3A}{\cos 3A}=3\tan 3A$
Hence, the value of tan A + tan (60 + A) – tan (60 – A) is 3 tan 3A.
So, option A is the correct answer.

Note: One can get confused with the formulae of sin 3A and cos 3A as mentioned in the solution.
Another approach for this question would be that we can use trigonometry identities;
$\begin{aligned} & \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \\\ & and \\\ & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \\\ \end{aligned}$
to simplify tan (60 + A) and tan (60 – A). Then simplify the results found to get the answer. Applying trigonometric identities always makes solutions flexible and shorter which is the key point of the solution.