Exercise 2.1

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Question 1

Find the principal value of

$$ \sin^{-1}\left(-\frac{1}{2}\right). $$

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Solution

The principal value of

$$ \sin^{-1}x $$

lies in the interval

$$ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. $$

Since

$$ \sin\left(-\frac{\pi}{6}\right)=-\frac12, $$

the required principal value is

$$ \boxed{-\frac{\pi}{6}}. $$

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Question 2

Find the principal value of

$$ \cos^{-1}\left(\frac{\sqrt3}{2}\right). $$

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Solution

The principal value of

$$ \cos^{-1}x $$

lies in the interval

$$ [0,\pi]. $$

Since

$$ \cos\left(\frac{\pi}{6}\right)=\frac{\sqrt3}{2}, $$

the required principal value is

$$ \boxed{\frac{\pi}{6}}. $$

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Question 3

Find the principal value of

$$ \operatorname{cosec}^{-1}(2). $$

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Solution

The principal value of

$$ \operatorname{cosec}^{-1}x $$

lies in the interval

$$ \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]. $$

Since

$$ \csc\left(\frac{\pi}{6}\right)=2, $$

the required principal value is

$$ \boxed{\frac{\pi}{6}}. $$

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Question 4

Find the principal value of

$$ \tan^{-1}\left(-\sqrt3\right). $$

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Solution

The principal value of

$$ \tan^{-1}x $$

lies in the interval

$$ \left(-\frac{\pi}{2},\frac{\pi}{2}\right). $$

Since

$$ \tan\left(-\frac{\pi}{3}\right)=-\sqrt3, $$

the required principal value is

$$ \boxed{-\frac{\pi}{3}}. $$

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Question 5

Find the principal value of

$$ \cos^{-1}\left(-\frac12\right). $$

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Solution

The principal value of

$$ \cos^{-1}x $$

lies in the interval

$$ [0,\pi]. $$

Since

$$ \cos\left(\frac{2\pi}{3}\right)=-\frac12, $$

the required principal value is

$$ \boxed{\frac{2\pi}{3}}. $$

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Question 6

Find the principal value of

$$ \tan^{-1}(-1). $$

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Solution

The principal value of

$$ \tan^{-1}x $$

lies in the interval

$$ \left(-\frac{\pi}{2},\frac{\pi}{2}\right). $$

Since

$$ \tan\left(-\frac{\pi}{4}\right)=-1, $$

the required principal value is

$$ \boxed{-\frac{\pi}{4}}. $$

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Question 7

Find the principal value of

$$ \sec^{-1}\left(\frac{2}{\sqrt3}\right). $$

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Solution

The principal value of

$$ \sec^{-1}x $$

lies in the interval

$$ [0,\pi]\setminus\left\{\frac{\pi}{2}\right\}. $$

Since

$$ \sec\left(\frac{\pi}{6}\right)=\frac{2}{\sqrt3}, $$

the required principal value is

$$ \boxed{\frac{\pi}{6}}. $$

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Question 8

Find the principal value of

$$ \cot^{-1}\left(\sqrt3\right). $$

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Solution

The principal value of

$$ \cot^{-1}x $$

lies in the interval

$$ (0,\pi). $$

Since

$$ \cot\left(\frac{\pi}{6}\right)=\sqrt3, $$

the required principal value is

$$ \boxed{\frac{\pi}{6}}. $$

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Question 9

Find the principal value of

$$ \cos^{-1}\left(-\frac{1}{\sqrt2}\right). $$

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Solution

The principal value of

$$ \cos^{-1}x $$

lies in the interval

$$ [0,\pi]. $$

Since

$$ \cos\left(\frac{3\pi}{4}\right)=-\frac{1}{\sqrt2}, $$

the required principal value is

$$ \boxed{\frac{3\pi}{4}}. $$

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Question 10

Find the principal value of

$$ \operatorname{cosec}^{-1}\left(-\sqrt2\right). $$

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Solution

The principal value of

$$ \operatorname{cosec}^{-1}x $$

lies in the interval

$$ \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]. $$

Since

$$ \csc\left(-\frac{\pi}{4}\right)=-\sqrt2, $$

the required principal value is

$$ \boxed{-\frac{\pi}{4}}. $$

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Question 11

Find the value of

$$ \tan^{-1}(1)+\cos^{-1}\left(-\frac12\right)+\sin^{-1}\left(-\frac12\right). $$

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Solution

Using the principal values,

$$ \tan^{-1}(1)=\frac{\pi}{4}, $$ $$ \cos^{-1}\left(-\frac12\right)=\frac{2\pi}{3}, $$

and

$$ \sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}. $$

Therefore,

$$ \begin{aligned} \tan^{-1}(1)+\cos^{-1}\left(-\frac12\right)+\sin^{-1}\left(-\frac12\right) &=\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}\\[2mm] &=\frac{\pi}{4}+\frac{\pi}{2}\\[2mm] &=\frac{3\pi}{4}. \end{aligned} $$

Answer:

$$ \boxed{\frac{3\pi}{4}} $$

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Question 12

Find the value of

$$ \cos^{-1}\left(\frac12\right)+2\sin^{-1}\left(\frac12\right). $$

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Solution

Using the principal values,

$$ \cos^{-1}\left(\frac12\right)=\frac{\pi}{3} $$

and

$$ \sin^{-1}\left(\frac12\right)=\frac{\pi}{6}. $$

Hence,

$$ \begin{aligned} \cos^{-1}\left(\frac12\right)+2\sin^{-1}\left(\frac12\right) &=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)\\[2mm] &=\frac{\pi}{3}+\frac{\pi}{3}\\[2mm] &=\frac{2\pi}{3}. \end{aligned} $$

Answer:

$$ \boxed{\frac{2\pi}{3}} $$

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Question 13

If

$$ \sin^{-1}x=y, $$

then choose the correct answer.

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Solution

The principal value of the inverse sine function always lies in the interval

$$ -\frac{\pi}{2}\le y\le\frac{\pi}{2}. $$

Hence, the correct option is

$$ \boxed{\text{Option (B)}} $$

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Question 14

Find the value of

$$ \tan^{-1}\sqrt3-\sec^{-1}(-2). $$

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Solution

Using the principal values,

$$ \tan^{-1}\sqrt3=\frac{\pi}{3} $$

and

$$ \sec^{-1}(-2)=\frac{2\pi}{3}. $$

Therefore,

$$ \begin{aligned} \tan^{-1}\sqrt3-\sec^{-1}(-2) &=\frac{\pi}{3}-\frac{2\pi}{3}\\[2mm] &=-\frac{\pi}{3}. \end{aligned} $$

Answer:

$$ \boxed{-\frac{\pi}{3}} $$

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