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Mathematics Question on Inverse Trigonometric Functions

Find the principal value tan−1(−3)+sec−1(−2)−cosec−1(23)tan^{-1}\left(-\sqrt{3}\right) + sec^{-1}\left(-2\right)-cosec^{-1}\left(\frac{2}{\sqrt{3}}\right)

A

5π6\frac{5\pi}{6}

B

2π3\frac{2\pi}{3}

C

π3\frac{\pi}{3}

D

00

Answer

00

Explanation

Solution

Let tan−1(−3)=αtan^{-1}\left(-\sqrt{3}\right) = \alpha ⇒tan α=−3=−tanπ3\Rightarrow tan \,\alpha = -\sqrt{3} = -tan \frac{\pi}{3} =tan(−π3)= tan\left(-\frac{\pi}{3}\right) ⇒ α=−π3∈(−π2,π2)\Rightarrow\, \alpha = \frac{-\pi }{3} \in \left( \frac{-\pi }{2}, \frac{\pi }{2}\right) ∴\therefore Principal value of tan−1(−3)tan^{-1} \left(-\sqrt{3}\right) is (−π3)\left(\frac{-\pi }{3}\right) Let sec−1(−2)=βsec^{-1} \left(-2\right) = \beta ⇒sec β=−2\Rightarrow sec\,\beta = -2 =−secπ3= -sec \frac{\pi}{3} =sec(π−π3)= sec\left(\pi-\frac{\pi }{3}\right) =sec(2π3)= sec \left(\frac{2\pi }{3}\right) \Rightarrow \beta = \frac{2\pi }{3} \in \left[0, \,\pi\right]-\left\\{\frac{\pi}{2}\right\\} ∴\therefore Principal value of sec−1(−2)sec^{-1} \left(-2\right) is 2π3\frac{2\pi }{3} Let cosec−1(23)=γcosec^{-1} \left(\frac{2}{\sqrt{3}}\right) = \gamma ⇒cosec γ=23\Rightarrow cosec\,\gamma = \frac{2}{\sqrt{3}} =cosecπ3= cosec \frac{\pi}{3} \Rightarrow \gamma = \frac{\pi }{3}\in\left[\frac{-\pi }{2}, \frac{\pi }{2}\right] - \left\\{0\right\\} ∴\therefore Principal value of cosec−1(23)cosec^{-1} \left(\frac{2}{\sqrt{3}}\right) is π3\frac{\pi }{3} So, the principal value of tan−1(−3)+sec−1(−2)−cosec−1(23)tan^{-1}\left(-\sqrt{3}\right) + sec^{-1}\left(-2\right)-cosec^{-1}\left(\frac{2}{\sqrt{3}}\right) =−π3+2π3−π3= \frac{-\pi }{3}+\frac{2\pi }{3} - \frac{\pi }{3} =0= 0