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

If cot⁡−1(x)+tan⁡−1(3)=π2{\cot ^{ - 1}}\left( x \right) + {\tan ^{ - 1}}\left( 3 \right) = \dfrac{\pi }{2}, then x=x =
A. 13\dfrac{1}{3}
B. 14\dfrac{1}{4}
C. 33
D. 44

Explanation

Solution

In order to solve the equation, initiate with proving the resultant value tan⁡−1x+cot⁡−1x=π2{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2} by following the appropriate steps then move ahead to compare the given values in the question with the obtained result. Solve the equations with the appropriate subtraction or addition method and get the value of xx.

Complete step by step answer:
Considering tan⁡−1x=p{\tan ^{ - 1}}x = p ……(1)
Multiplying both the sides by tan⁡\tan , we get:
tan⁡(tan⁡−1x)=tan⁡p\tan \left( {{{\tan }^{ - 1}}x} \right) = \tan p
⇒x=tan⁡p\Rightarrow x = \tan p
Since, we know that tan⁡p\tan p can be written in terms of cot⁡\cot as tan⁡p=cot⁡(π2−p)\tan p = \cot \left( {\dfrac{\pi }{2} - p} \right).So, substituting this value in the above function, we get:
⇒x=cot⁡(π2−p)\Rightarrow x = \cot \left( {\dfrac{\pi }{2} - p} \right)
Multiplying both the sides by cot⁡−1{\cot ^{ - 1}}, we get:
⇒cot⁡−1x=cot⁡−1(cot⁡(π2−p))\Rightarrow {\cot ^{ - 1}}x = {\cot ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{2} - p} \right)} \right)
⇒cot⁡−1x=π2−p\Rightarrow {\cot ^{ - 1}}x = \dfrac{\pi }{2} - p ……(2)

Adding equation 1 and 2 we get:
tan⁡−1x+cot⁡−1x=p+π2−p{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = p + \dfrac{\pi }{2} - p
Solving it further, we get:
⇒tan⁡−1x+cot⁡−1x=π2\Rightarrow {\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2} …..(3)
Subtracting both sides by cot⁡−1(x){\cot ^{ - 1}}\left( x \right):
⇒tan⁡−1x+cot⁡−1x−cot⁡−1x=π2−cot⁡−1x\Rightarrow {\tan ^{ - 1}}x + {\cot ^{ - 1}}x - {\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\cot ^{ - 1}}x
⇒tan⁡−1x=π2−cot⁡−1x\Rightarrow {\tan ^{ - 1}}x = \dfrac{\pi }{2} - {\cot ^{ - 1}}x ……(4)
Now, since we are given with the equation cot⁡−1(x)+tan⁡−1(3)=π2{\cot ^{ - 1}}\left( x \right) + {\tan ^{ - 1}}\left( 3 \right) = \dfrac{\pi }{2}.
Subtracting both the sides by cot⁡−1(x){\cot ^{ - 1}}\left( x \right),we get:
cot⁡−1(x)+tan⁡−1(3)−cot⁡−1(x)=π2−cot⁡−1(x){\cot ^{ - 1}}\left( x \right) + {\tan ^{ - 1}}\left( 3 \right) - {\cot ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} - {\cot ^{ - 1}}\left( x \right)
⇒tan⁡−1(3)=π2−cot⁡−1(x)\Rightarrow {\tan ^{ - 1}}\left( 3 \right) = \dfrac{\pi }{2} - {\cot ^{ - 1}}\left( x \right)
From equation 4, we can write it as:
⇒tan⁡−1(3)=tan⁡−1(x)\Rightarrow {\tan ^{ - 1}}\left( 3 \right) = {\tan ^{ - 1}}\left( x \right)
Multiplying both the sides by tan⁡\tan , we get:
⇒tan⁡(tan⁡−1(3))=tan⁡(tan⁡−1(x))\Rightarrow \tan \left( {{{\tan }^{ - 1}}\left( 3 \right)} \right) = \tan \left( {{{\tan }^{ - 1}}\left( x \right)} \right)
⇒3=x\Rightarrow 3 = x
∴x=3\therefore x = 3
Therefore, the value of x=3x = 3.

Hence, option C is correct.

Note: We could have alternatively directly solved the equation by comparing the given equation with the obtained equation 3, which would have resulted in the value of xx to be 33. It’s important to prove the results before applying direct values otherwise it may lead to errors. tan⁡p\tan p is written in terms of cot⁡\cot as tan⁡p=cot⁡(π2−p)\tan p = \cot \left( {\dfrac{\pi }{2} - p} \right) because π2\dfrac{\pi }{2} is an odd function, so it changes the value outside the bracket into its reverse one and obtains a positive value because π2−p\dfrac{\pi }{2} - p lies in the first quadrant and in first quadrant all the values are positive. Similarly, we can write cot in terms of tan.