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Question

Mathematics Question on Inverse Trigonometric Functions

If tan⁡(x+y)=33\tan(x + y) = 33 and x=tan⁡−1 3,x = \tan^{-1} \, 3, then yy is

A

310\frac{3}{10}

B

3310\frac{33}{10}

C

13\frac{1}{3}

D

tan⁡−1310\tan^{-1} \frac{3}{10}

Answer

tan⁡−1310\tan^{-1} \frac{3}{10}

Explanation

Solution

We have, tan⁡(x+y)=33\tan(x + y) = 33
⇒  tan⁡ x+tan⁡ y−tan⁡ x tan⁡ y=33\Rightarrow \:\: \frac{\tan \, x + \tan \, y}{- \tan \, x \, \tan \, y} = 33
⇒  3+tan⁡ y1−3 tan⁡ y=33    (∵ x=tan⁡−13 ⇒tan⁡ x=3)\Rightarrow \:\: \frac{3 + \tan \, y}{1 - 3 \, \tan \, y} = 33 \:\:\:\: (\because \, x = \tan^{-1} 3 \: \Rightarrow \tan \, x = 3)
⇒  3+tan⁡ y=33−99 tan⁡ y\Rightarrow \:\: 3 + \tan \, y = 33 - 99 \, \tan \, y
⇒  100 tan⁡ y=30\Rightarrow \:\: 100 \, \tan \, y = 30
⇒  tan⁡ y=310 ⇒ y=tan⁡−1310\Rightarrow \:\: \tan \, y = \frac{3}{10} \: \Rightarrow \: y = \tan^{-1} \frac{3}{10}