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Chapter 2: Problem 14

Find the values of the following: \(\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)\) is equal to (A) \(\pi\) (B) \(-\frac{\pi}{3}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{2 \pi}{3}\)

Short Answer

Expert verified
\(-\frac{\pi}{3}\)

Step by step solution

01

Find the value of \(\tan^{-1}\sqrt{3}\)

The value of \(\tan^{-1}\sqrt{3}\) corresponds to the angle whose tangent is \(\sqrt{3}\). This angle is \(\frac{\pi}{3}\) because \(\tan(\frac{\pi}{3}) = \sqrt{3}\).
02

Find the value of \(\sec^{-1}(-2)\)

The value of \(\sec^{-1}(-2)\) corresponds to the angle whose secant is -2. In the interval \( [0, \pi] \) where the \(\sec^{-1}\) function is defined, the corresponding angle is \(\pi - \frac{\pi}{3} = \frac{2\pi}{3}\) because \(\sec(\frac{2\pi}{3}) = -2\).
03

Evaluate the difference

Subtract the value found in step 2 from the value found in step 1: \(\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}\). The result is option (B).

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Key Concepts

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tan inverse
here is a great resource that delves into constructions based on inverse trigonometric functions.

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Most popular questions from this chapter

Find the principal values of the following: $$ \operatorname{cosec}^{-1}(2) $$

Find the values of each of the following: $$ \tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right],|x|<1, y>0 \text { and } x y<1 $$

$$ 2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7} $$

$$ \cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65} $$

Find the principal values of the following: $$ \tan ^{-1}(-\sqrt{3}) $$
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