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Question

Mathematics Question on Inverse Trigonometric Functions

Find the value of tan−13−sec−1(−2)tan^{-1\sqrt3-sec^{-1}(-2)} is equal to

A

π\pi

B

−π3-\frac {\pi}{3}

C

π3\frac {\pi}{3}

D

2π3\frac{2\pi}{3}

Answer

−π3-\frac {\pi}{3}

Explanation

Solution

Let tan−13=x.tan^{-1\sqrt3=x}.
Then tan x=3=tanπ3tan\,x=\sqrt3=tan\frac{\pi}{3}.
We know that the range of principal vale branch of tan−1 is(−π2,π2)tan^{-1}\,is \bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg)
therefore tan−13=π3tan^{-1}\sqrt3=\frac{\pi}{3}
Let sec−1(2)=y.sec^{-1}(2)=y.
Then, sec y=−2=−sec(π3)=sec(π−π3)=sec2π3.sec \,y=-2=-sec(\frac{\pi}{3})=sec (\pi-\frac{\pi}{3})=sec\frac{2\pi}{3}.
We know that the range of principal vale branch of sec-1 is [0,π]−π2.[0,\pi]-\\{\frac{\pi}{2}\\}.
sec−1(−2)=2π3.sec^{-1}(-2)=\frac{2\pi}{3}.
Hence, tan−1(3)−sec−1(2)=π3−2π3=−π3tan^{-1}(\sqrt3)-sec^{-1}(2)=\frac{\pi}{3}-\frac{2\pi}{3}=-\frac{\pi}{3}