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Question: If \(\cot ^ { - 1 } x + \tan ^ { - 1 } 3 = \frac { \pi } { 2 }\), then x =...

If cot⁡−1x+tan⁡−13=π2\cot ^ { - 1 } x + \tan ^ { - 1 } 3 = \frac { \pi } { 2 }, then x =

A

1/3

B

¼

C

3

D

4

Answer

3

Explanation

Solution

We have cot⁡−1x+tan⁡−13=π2\cot ^ { - 1 } x + \tan ^ { - 1 } 3 = \frac { \pi } { 2 }

⇒cot⁡−1x+tan⁡−13=π2⇒tan⁡−11x+tan⁡−13=π2\Rightarrow \cot ^ { - 1 } x + \tan ^ { - 1 } 3 = \frac { \pi } { 2 } \Rightarrow \tan ^ { - 1 } \frac { 1 } { x } + \tan ^ { - 1 } 3 = \frac { \pi } { 2 }

⇒tan⁡−1(1x+31−1x⋅3)=tan⁡−1(10)\Rightarrow \tan ^ { - 1 } \left( \frac { \frac { 1 } { x } + 3 } { 1 - \frac { 1 } { x } \cdot 3 } \right) = \tan ^ { - 1 } \left( \frac { 1 } { 0 } \right)

⇒3x+1x−3=10⇒x=3\Rightarrow \frac { 3 x + 1 } { x - 3 } = \frac { 1 } { 0 } \Rightarrow x = 3

Aliter : As we know that, tan⁡−1x+cot⁡−1x=π2\tan ^ { - 1 } x + \cot ^ { - 1 } x = \frac { \pi } { 2 } therefore for the given question, x should be 3.