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Question

Mathematics Question on Inverse Trigonometric Functions

In a ΔABC\Delta ABC, if A=tan−1 2A = tan^{-1}\, 2 and B=tan−1 3B = tan^{ -1}\, 3 , then C=C =

A

π3\frac{\pi}{3}

B

π4\frac{\pi}{4}

C

π6\frac{\pi}{6}

D

3π4\frac{3\pi}{4}

Answer

π4\frac{\pi}{4}

Explanation

Solution

We have, A=tan−1 2⇒tan A=2A=tan^{-1}\,2 \Rightarrow tan\,A=2 and B=tan−1 3⇒tan B=3B=tan^{-1}\,3 \Rightarrow tan\,B=3 since A,B,CA, B, C are angles of a triangle, A+B+C=πA+B+C = \pi ⇒C=π−(A+B)\Rightarrow C=\pi-\left(A+B\right) Now, A+B=tan−1 2+tan−1 3=π+tan−1[2+31−2.3]A+B=tan^{-1}\,2+tan^{-1}\,3=\pi+tan^{-1}\left[\frac{2+3}{1-2.3}\right] [∴tan−1 x+tan−1 y=π+tan−1[x+y1−xy]for x>0,y>0 and xy>1]\left[\therefore tan^{-1}\,x+tan^{-1}\,y=\pi+tan^{-1}\left[\frac{x+y}{1-xy}\right] for \,x>0, y>0\,and\,xy>1\right] =π+tan−1(−1)=π−tan−1 1=π−π4=3π4=\pi+tan^{-1}\left(-1\right)=\pi-tan^{-1}\,1=\pi-\frac{\pi}{4}=\frac{3\pi}{4} ∴From(1),C=π−3π4=π4\therefore From \left(1\right), C=\pi-\frac{3\pi}{4}=\frac{\pi}{4}