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Question: If \[ta{{n}^{-1}}a+ta{{n}^{-1}}b+ta{{n}^{-1}}c=\pi \], prove that a+ b + c = abc....

If tan−1a+tan−1b+tan−1c=πta{{n}^{-1}}a+ta{{n}^{-1}}b+ta{{n}^{-1}}c=\pi , prove that a+ b + c = abc.

Explanation

Solution

Hint: For solving this problem, first we apply the addition identity for the inverse of tan function to simplify the expression on the left hand side. After getting a simplified expression, we apply tan function to both sides. By using this methodology, we can easily obtain our answer.

Complete step-by-step answer:
First, we take left-hand side to simplify:
⇒tan−1a+tan−1b+tan−1c\Rightarrow ta{{n}^{-1}}a+ta{{n}^{-1}}b+ta{{n}^{-1}}c
Now, by applying the trigonometric property tan−1x+tan−1y=tan⁡−1(x+y1−xy)ta{{n}^{-1}}x+ta{{n}^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right), we get
⇒tan−1(a+b1−ab)+tan−1c\Rightarrow ta{{n}^{-1}}\left( \dfrac{a+b}{1-ab} \right)+ta{{n}^{-1}}c
Applying the same formula again in the tan−1(a+b1−ab)+tan−1cta{{n}^{-1}}\left( \dfrac{a+b}{1-ab} \right)+ta{{n}^{-1}}c to simplify it further, we get
⇒tan−1(a+b1−ab+c1−(a+b1−ab)c) ⇒tan−1(a+b+c−abc1−ab1−ab−ac−bc1−ab) ⇒tan−1(a+b+c−abc1−ab−bc−ca) \begin{aligned} & \Rightarrow ta{{n}^{-1}}\left( \dfrac{\dfrac{a+b}{1-ab}+c}{1-\left( \dfrac{a+b}{1-ab} \right)c} \right) \\\ & \Rightarrow ta{{n}^{-1}}\left( \dfrac{\dfrac{a+b+c-abc}{1-ab}}{\dfrac{1-ab-ac-bc}{1-ab}} \right) \\\ & \Rightarrow ta{{n}^{-1}}\left( \dfrac{a+b+c-abc}{1-ab-bc-ca} \right) \\\ \end{aligned}
Now, by using the right-hand side, we get
tan⁡−1(a+b+c−abc1−ab−bc−ca)=π{{\tan }^{-1}}\left( \dfrac{a+b+c-abc}{1-ab-bc-ca} \right)=\pi
Applying tan operation to both sides, we get
a+b+c−abc1−ab−bc−ca=tan⁡(π)\dfrac{a+b+c-abc}{1-ab-bc-ca}=\tan (\pi )
As we know that the value of tan⁡(π)\tan (\pi ) is 0 and tan⁡−1(tan⁡θ)=θ{{\tan }^{-1}}\left( \tan \theta \right)=\theta , so we simplified the above expression as,
a+b+c−abc1−ab−bc−ca=0\dfrac{a+b+c-abc}{1-ab-bc-ca}=0
Now, taking the denominator of the left-hand side to the right-hand side and multiplying it with 0. So, the right-hand side is 0. So, the remaining equation is
⇒a+b+c−abc=0 ⇒a+b+c=abc \begin{aligned} & \Rightarrow a+b+c-abc=0 \\\ & \Rightarrow a+b+c=abc \\\ \end{aligned}
Hence, we proved the equivalence of both sides as required in the problem.

Note: The key concept involved in solving this problem is the knowledge of inverse properties related to tan function. Students must be careful while solving the fraction which involves a variety of terms. Silly mistakes are bound to occur in that particular section.