Question

The value of $\tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right]$ is equal to:
(A) $\dfrac{5}{6}$
(B) $\dfrac{7}{6}$
(C) $\dfrac{1}{6}$
(D) $\dfrac{1}{7}$

Answer 1

Hint: The given question deals with finding the value of trigonometric expression doing basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$ and double angle formula for cosine. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We must know the simplification rules to solve the problem with ease.

Complete step-by-step solution:
In the given problem, we have to find value of $\tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right]$. So, the angle given in the trigonometric function tangent is the difference of two angles whose tangent is already given to in the problem.
So, we have, $\tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right]$
Expanding the trigonometric function using the compound angle formula for tangent $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$. So, we get,
$ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right] = \dfrac{{\tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right) - \tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)}}{{1 + \tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)\tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)}}$
Now, we know that $\tan \left( {{{\tan }^{ - 1}}x} \right) = x$. So, we get,
$ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right] = \dfrac{{\left( {\dfrac{1}{2}} \right) - \left( {\dfrac{1}{3}} \right)}}{{1 + \left( {\dfrac{1}{2}} \right) \times \left( {\dfrac{1}{3}} \right)}}$
Taking the LCM in numerator, we get,
$ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right] = \dfrac{{\dfrac{{3 - 2}}{6}}}{{1 + \left( {\dfrac{1}{2}} \right) \times \left( {\dfrac{1}{3}} \right)}}$
Simplifying the calculations, we get,
$ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right] = \dfrac{{\dfrac{{3 - 2}}{6}}}{{1 + \dfrac{1}{6}}}$
$ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right] = \dfrac{{\left( {\dfrac{1}{6}} \right)}}{{\left( {\dfrac{7}{6}} \right)}}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right] = \dfrac{1}{7}$
Therefore, value of $\tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right]$ is $\dfrac{1}{7}$.
Hence, Option (D) is the correct answer.

Note: We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. We should take care of the calculations while solving such problems.