Question

Find the value of \[\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)\]
A) \[\dfrac{1}{\sqrt{2}}\]
B) \[\dfrac{\sqrt{3}}{2}\]
C) \[\dfrac{1}{2}\]
D) \[\dfrac{\pi }{4}\]

Answer 1

Hint: In this particular problem you have to use the formula for \[{{\tan }^{-1}}\left( A \right)+{{\tan }^{-1}}\left( B \right)={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\], Comparing with the given problem we have the value of A and B. We simplify this by the help of LCM method and on further we use the standard trigonometric values to get the desired results.

Complete step by step answer:
In this type of problems, expression is given that \[\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)---(1)\]
We have to find the value of this above equation for that we need to use the trigonometry formula that is \[{{\tan }^{-1}}\left( A \right)+{{\tan }^{-1}}\left( B \right)={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\]
Here, in this problem A and B is given in the equation (1) that is \[A=\dfrac{1}{3}\]and\[B=\dfrac{1}{2}\]substitute this value in above formula we get:
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1-\left( \dfrac{1}{3} \right)\left( \dfrac{1}{2} \right)} \right)\]
Now we have to simplify the above equation by taking LCM we get:
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{5}{6}}{1-\left( \dfrac{1}{6} \right)} \right)\]
Multiply the above the equation by 6 on numerator as well as denominator
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{5}{6}\times 6}{6\times \left[ 1-\left( \dfrac{1}{6} \right) \right]} \right)\]
By further simplification and further solving this we get:
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( \dfrac{5}{6-1} \right)\]
By further solving this above equation we get:
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( \dfrac{5}{5} \right)\]
If you observe this above equation then you can see that 5 get cancelled that means it remains 1 that is \[{{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}\]
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( 1 \right)\]
As we know that \[{{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}\]substitute this value in above equation we get:
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{4}---(2)\]
As we get the equation (2)
Now, we have to substitute the value of equation (2) on equation (1) we get:
\[\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)=\cos \left( \dfrac{\pi }{4} \right)\]
As we already know that\[\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\]and substitute this value in above equation we get:
\[\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)=\dfrac{1}{\sqrt{2}}\]
Therefore, the correct option is option (A).

Note:
In this problem we have to keep in mind which formula is to be used for particular problems. So, read the question carefully about what is given and what we have to find based on that we have to use the formula. In this case we have to use the trigonometry property. Remember all the trigonometric values such as \[\tan \left( \dfrac{\pi }{4} \right)=1\].