Question

Prove the given inverse trigonometric equation as : ${{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18={{\cot }^{-1}}3$.

Answer 1

Hint: Change the given cot inverse functions into tan inverse functions by using the formula: ${{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}$, for ‘x’ greater than 0, so that we have to prove: ${{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}+{{\tan }^{-1}}\dfrac{1}{18}={{\tan }^{-1}}\dfrac{1}{3}$. Now, take the sum of first two terms and use the formula: ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$, where $ab$ must be less than 1, to simplify. Then take the sum of this obtained expression with the third term. Use the same formula: ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$, to get the answer.

Complete step-by-step solution -
We have to prove: ${{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18={{\cot }^{-1}}3$
Converting these cot inverse functions into tan inverse functions by using the formula: ${{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}$, for ‘x’ greater than 0, we have to prove: ${{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}+{{\tan }^{-1}}\dfrac{1}{18}={{\tan }^{-1}}\dfrac{1}{3}$.
Now, considering the sum ${{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}$ by using the identity: ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$, where $ab$ must be less than 1, we get,
$\dfrac{1}{7}\times \dfrac{1}{8}=\dfrac{1}{56}$, which is less than 1. Therefore,
${{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{7}+\dfrac{1}{8}}{1-\dfrac{1}{7}\times \dfrac{1}{8}} \right)$
Taking L.C.M and simplifying, we get,
$\begin{align}
  & {{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}={{\tan }^{-1}}\left( \dfrac{\dfrac{8+7}{56}}{1-\dfrac{1}{56}} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\dfrac{15}{56}}{\dfrac{56-1}{56}} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\dfrac{15}{56}}{\dfrac{55}{56}} \right) \\
\end{align}$
Cancelling the common terms, we get,
${{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}={{\tan }^{-1}}\dfrac{3}{11}$
Now considering the sum ${{\tan }^{-1}}\dfrac{3}{11}+{{\tan }^{-1}}\dfrac{1}{18}$, where $\dfrac{3}{11}\times \dfrac{1}{18}=\dfrac{1}{66}$ is less than 1, we get,
${{\tan }^{-1}}\dfrac{3}{11}+{{\tan }^{-1}}\dfrac{1}{18}={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{11}+\dfrac{1}{18}}{1-\dfrac{3}{11}\times \dfrac{1}{18}} \right)$
Taking L.C.M and simplifying, we get,
$\begin{align}
  & {{\tan }^{-1}}\dfrac{3}{11}+{{\tan }^{-1}}\dfrac{1}{18}={{\tan }^{-1}}\left( \dfrac{\dfrac{18\times 3+11}{18\times 11}}{1-\dfrac{3}{18\times 11}} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\dfrac{54+11}{18\times 11}}{\dfrac{18\times 11-3}{18\times 11}} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{65}{195} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{1}{3} \right) \\
 & =R.H.S \\
\end{align}$
Therefore, it is proved that: ${{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}+{{\tan }^{-1}}\dfrac{1}{18}={{\tan }^{-1}}\dfrac{1}{3}$.
Hence, ${{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18={{\cot }^{-1}}3$.

Note: One may note that we have changed the given cot inverse functions into tan inverse functions because generally, we remember the formula of the sum of two tan inverse functions and not cot inverse functions. You can remember the formula for cot inverse functions for solving the question in fewer steps. Remember that in the above question, it is difficult to solve while taking the sum of all the three terms together. Therefore, we have considered the sum of two terms at a time.