Solveeit Logo
Login

Question

Question: If \({\tan ^{ - 1}}x + {\tan ^{ - 1}}3 = {\tan ^{ - 1}}8\) then x=? \( {\text{A}}{\text{. }}\dfr...

If tan⁡−1x+tan⁡−13=tan⁡−18{\tan ^{ - 1}}x + {\tan ^{ - 1}}3 = {\tan ^{ - 1}}8 then x=?
A. 13 B. 15 C. 3 D. 5  {\text{A}}{\text{. }}\dfrac{1}{3} \\\ {\text{B}}{\text{. }}\dfrac{1}{5} \\\ {\text{C}}{\text{. 3}} \\\ {\text{D}}{\text{. 5}} \\\

Explanation

Solution

Hint: -To solve this question first we use formula of inverse trigonometric function tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) and then proceed further to find the value of x .

Complete step-by-step answer:
We have
tan⁡−1x+tan⁡−13=tan⁡−18{\tan ^{ - 1}}x + {\tan ^{ - 1}}3 = {\tan ^{ - 1}}8
Now as given in hint use the formula tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)
We get,
tan⁡−1(3+x1−3x){\tan ^{ - 1}}\left( {\dfrac{{3 + x}}{{1 - 3x}}} \right) = tan⁡−18{\tan ^{ - 1}}8
Now we can cancel out tan⁡−1{\tan ^{ - 1}} because it is common on both sides to equal sign.
After cancel out we get,
(3+x1−3x)\left( {\dfrac{{3 + x}}{{1 - 3x}}} \right) =88
On cross multiplying we get,
3+x = 8- 24x
25x = 5
∴x=15\therefore x = \dfrac{1}{5}
Hence option B is the correct option.

Note: - Whenever we get this type of question the key concept of solving is we have to follow the standard formula of addition of inverse of tan then simple calculation. One thing we should make sure of in questions of inverse trigonometric function is that we have to care about domain and range so that we can check if our answer is correct or not and many questions come on the domain and range of inverse trigonometric functions.