Question

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

Answer 1

We will use the property of inverse trigonometric function that ${{\cot }^{-1}}x={{\tan }^{-1}}\left( \dfrac{1}{x} \right)$, if $x>0$ and then we will use the formula as follows:
${{\tan }^{-1}}x+{{\tan }^{1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{a-xy} \right)$
So by using this, we will get the required value of the expression.

Complete step-by-step answer:
We have been asked to prove that ${{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18={{\cot }^{-1}}3$.
We know the property of inverse trigonometric function that ${{\cot }^{-1}}x={{\tan }^{-1}}\left( \dfrac{1}{x} \right)$, is $x>0$
Since we already know that 7>0, we can write the condition as
${{\cot }^{-1}}7={{\tan }^{-1}}\left( \dfrac{1}{7} \right)......(1)$
Since we already know that 8>0, we can write the condition as
${{\cot }^{-1}}8={{\tan }^{-1}}\left( \dfrac{1}{8} \right).......(2)$
And since we already know that 18>0, again we can write the condition as
${{\cot }^{-1}}18={{\tan }^{-1}}\left( \dfrac{1}{18} \right).......(3)$
On adding equations (1), (2) and (3), we get as follows:
$\Rightarrow {{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18={{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}+{{\tan }^{-1}}\dfrac{1}{18}.......(4)$
Now, applying the formula of trigonometric function given as follows in equation (4):
${{\tan }^{-1}}x+ta{{n}^{-1}}y=ta{{n}^{-1}}\left( \dfrac{x+y}{1-xy} \right)$, if $xy<1$
In equation (4), we get as follows:
$\Rightarrow {{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18={{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}+{{\tan }^{-1}}\dfrac{1}{18}$
Solving the RHS of the above equation further, we get as follows:

& \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{1}{7}+\dfrac{1}{8}}{1-\dfrac{1}{7}\times \dfrac{1}{8}} \right)+{{\tan }^{-1}}\dfrac{1}{18} \\\ & ={{\tan }^{-1}}\left( \dfrac{\dfrac{15}{56}}{\dfrac{56-1}{56}} \right)+{{\tan }^{-1}}\dfrac{1}{18} \\\ & ={{\tan }^{-1}}\left( \dfrac{15}{55} \right)+{{\tan }^{-1}}\left( \dfrac{1}{18} \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{3}{11} \right)+{{\tan }^{-1}}\left( \dfrac{1}{18} \right) \\\ \end{aligned}$$ Hence we get $${{\tan }^{-1}}\left( \dfrac{3}{11} \right)+{{\tan }^{-1}}\left( \dfrac{1}{18} \right)........(5)$$ Again, applying the formula $${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$ in equation (5) we get as follows: $$\begin{aligned} & \Rightarrow {{\tan }^{-1}}\dfrac{1}{7}+{{\tan }^{-1}}\dfrac{1}{8}+{{\tan }^{-1}}\dfrac{1}{18} \\\ & ={{\tan }^{-1}}\left( \dfrac{3}{11} \right)+{{\tan }^{-1}}\left( \dfrac{1}{18} \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{11}+\dfrac{1}{18}}{1-\dfrac{3}{11}\times \dfrac{1}{18}} \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{\dfrac{54+11}{198}}{\dfrac{198-3}{198}} \right)={{\tan }^{-1}}\left( \dfrac{65}{195} \right)={{\tan }^{-1}}\left( \dfrac{1}{3} \right) \\\ \end{aligned}$$ Hence, we get $${{\tan }^{-1}}\left( \dfrac{1}{3} \right)$$. Since we already know that $${{\tan }^{-1}}\dfrac{1}{x}={{\cot }^{-1}}x$$, so if x>1 then $${{\cot }^{-1}}(3)=$$ right hand side of the equation Hence the given expression is proved. **Note:** Be careful while doing calculation and also take care of the sign during the calculation. Also, remember that the condition to use the formula or property of inverse trigonometry is very important since the formula changes according to the condition so you make check the given condition before using the property or formula of the inverse trigonometric function.