Question

Prove the result of the following expression, $2{{\tan }^{-1}}\dfrac{3}{4}-{{\tan }^{-1}}\dfrac{17}{31}=\dfrac{\pi }{4}$.

Answer 1

Hint:To prove the result of the given expression, we should know a few inverse trigonometric formulas like, $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$. We also have to remember that $\tan \dfrac{\pi }{4}=1$. We can prove the given equality by using these formulas and should be very careful while doing the calculations.

Complete step-by-step answer:
In this question, we have been asked to prove the equality, $2{{\tan }^{-1}}\dfrac{3}{4}-{{\tan }^{-1}}\dfrac{17}{31}=\dfrac{\pi }{4}$. So, to prove this equality, we will first consider the left hand side of the equation. So, we can write it as,
$LHS=2{{\tan }^{-1}}\dfrac{3}{4}-{{\tan }^{-1}}\dfrac{17}{31}$
Now, we know that $2{{\tan }^{-1}}x$ can be expressed as ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. Hence, applying it to the expression, we get the LHS as,
$LHS={{\tan }^{-1}}\left[ \dfrac{2\left( \dfrac{3}{4} \right)}{1-{{\left( \dfrac{3}{4} \right)}^{2}}} \right]-{{\tan }^{-1}}\dfrac{17}{31}$
Now, we will simplify it further and get the LHS as,
$\begin{aligned} & LHS={{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{3}{2} \right)}{1-\left( \dfrac{9}{16} \right)} \right]-{{\tan }^{-1}}\dfrac{17}{31} \\\ & \Rightarrow LHS={{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{3}{2} \right)}{\dfrac{\left( 16-9 \right)}{16}} \right]-{{\tan }^{-1}}\dfrac{17}{31} \\\ & \Rightarrow LHS={{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{3}{2} \right)}{\left( \dfrac{7}{16} \right)} \right]-{{\tan }^{-1}}\dfrac{17}{31} \\\ \end{aligned}$
And we can further simplify it as,
$\begin{aligned} & LHS={{\tan }^{-1}}\left[ \dfrac{3\times 16}{2\times 7} \right]-{{\tan }^{-1}}\dfrac{17}{31} \\\ & \Rightarrow LHS={{\tan }^{-1}}\left[ \dfrac{24}{7} \right]-{{\tan }^{-1}}\dfrac{17}{31} \\\ \end{aligned}$
Now, we know that ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$, so for $a=\dfrac{24}{7}$ and $b=\dfrac{17}{31}$, we can apply this and write the above expression as,
$LHS={{\tan }^{-1}}\left[ \dfrac{\dfrac{24}{7}-\dfrac{17}{31}}{1+\dfrac{24}{7}\times \dfrac{17}{31}} \right]$
Now, we will take the LCM of both the terms of the numerator and the denominator. So, we will get the LHS as follows,
$LHS={{\tan }^{-1}}\left[ \dfrac{\dfrac{24\times 31-17\times 7}{7\times 31}}{\dfrac{7\times 31+24\times 17}{7\times 31}} \right]$
We can further write it as,
$\begin{aligned} & LHS={{\tan }^{-1}}\left[ \dfrac{\left( 24\times 31-17\times 7 \right)\left( 7\times 31 \right)}{\left( 7\times 31+24\times 17 \right)\left( 7\times 31 \right)} \right] \\\ & \Rightarrow LHS={{\tan }^{-1}}\left( \dfrac{744-119}{217+408} \right) \\\ & \Rightarrow LHS={{\tan }^{-1}}\left( \dfrac{625}{625} \right) \\\ \end{aligned}$
And we know that the common terms of the numerator and the denominator gets cancelled, so we can write the above as,
$LHS={{\tan }^{-1}}1$
Now, we know that $\tan \dfrac{\pi }{4}=1$. So, we can write $\dfrac{\pi }{4}={{\tan }^{-1}}1$. So, applying that, we can write the LHS as, $LHS=\dfrac{\pi }{4}$, which implies that it is equal to the right hand side or the RHS, given in the question. So, LHS = RHS.
Hence, we have proved the given expression.

Note: While solving the question, there are high possibilities that we end up with $LHS\ne RHS$, which would mean that there is a calculation mistake somewhere and we need to check the calculations again. Also, we must remember that, $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$ to solve this question easily.