Question

Find the value of the expression,\[{{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( \dfrac{-1}{7} \right)\].

Answer 1

Hint: In this question, we have two inverse tan functions in LHS and we have to find its value after solving. Here, we should not think to switch in any other inverse trigonometric function. Solve this equation in inverse tan function only. For that, we have already a formula of inverse tan, \[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)\] . We consider x as \[\dfrac{3}{4}\] and y as \[\dfrac{-1}{7}\] . Put these in the formula and solve it further.

Complete step-by-step solution:
We have to use a formula for solving the given question.
\[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)\]………..(1)
Replacing x by \[\dfrac{3}{4}\] and y by \[\dfrac{\left( -1 \right)}{\left( 7 \right)}\] in equation (1), we get
\[{{\tan }^{-1}}\dfrac{3}{4}-{{\tan }^{-1}}\left( \dfrac{-1}{7} \right)\]…………………(2)
Solving equation (2), using RHS of equation (1), we get
\[{{\tan }^{-1}}\left[ \dfrac{\dfrac{3}{4}-\dfrac{-1}{7}}{1+\left( \dfrac{3}{4} \right)\left( \dfrac{-1}{7} \right)} \right]\]
Taking \[4\times \left( 7 \right)\] as LCM in numerator and denominator, we get
\[{{\tan }^{-1}}\left( \dfrac{\dfrac{3(7)-4(-1)}{4\times(7)}}{\dfrac{4(7)+3(-1)}{4\times (7)}} \right)\]
The term \[4\times \left( 7 \right)\] , is present both in numerator and denominator. So, we can cancel \[4\times \left( 7 \right)\] . After canceling the term \[4\times \left( 7 \right)\] , our equation looks like
\[{{\tan }^{-1}}\left( \dfrac{21+4}{21+4} \right)\]
\[={{\tan }^{-1}}\left( \dfrac{25}{25} \right)\]
The term \[25\] is present both in the numerator as well as the denominator.
So, dividing by \[\left( 25 \right)\] in both numerator and denominator, we get
\[{{\tan }^{-1}}(1)\]…………(3)
We also know that, \[\tan \left( \dfrac{\pi }{4} \right)=1\]…………..(4)
Substituting equation (3) in equation (4), we get
\[{{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)\]
\[=\dfrac{\pi }{4}\]
Therefore,\[{{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( \dfrac{-1}{7} \right)=\dfrac{\pi }{4}\].

Note: In this question, one can think to transform the inverse tan function to any other trigonometric form. But it will only increase the complexity of the solution. On the other hand, someone can think of finding the principal values of both inverse tan functions in the LHS of the expression. But the restriction on values of x and y is not given in the question. So, we can’t continue with this approach.