Miscellaneous Exercise

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Question 1

Find the value of

$$ \cos^{-1}\left(\cos\frac{13\pi}{6}\right). $$

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Solution

The principal value of

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

lies in the interval

$$ [0,\pi]. $$

Now,

$$ \cos\frac{13\pi}{6} = \cos\left(2\pi+\frac{\pi}{6}\right) = \cos\frac{\pi}{6} = \frac{\sqrt3}{2}. $$

Hence,

$$ \cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \cos^{-1}\left(\frac{\sqrt3}{2}\right). $$

Since

$$ \cos\frac{\pi}{6} = \frac{\sqrt3}{2}, $$

and

$$ \frac{\pi}{6}\in[0,\pi], $$

the required principal value is

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

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Question 2

Find the value of

$$ \tan^{-1}\left(\tan\frac{7\pi}{6}\right). $$

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Solution

The principal value of

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

lies in the interval

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

Now,

$$ \tan\frac{7\pi}{6} = \tan\left(\pi+\frac{\pi}{6}\right) = \tan\frac{\pi}{6} = \frac{1}{\sqrt3}. $$

Hence,

$$ \tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \tan^{-1}\left(\frac{1}{\sqrt3}\right). $$

Since

$$ \tan\frac{\pi}{6} = \frac{1}{\sqrt3}, $$

and

$$ \frac{\pi}{6}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), $$

the required principal value is

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

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Question 3

Prove that

$$ 2\sin^{-1}\left(\frac35\right) = \tan^{-1}\left(\frac{24}{7}\right). $$

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Proof

Let

$$ \theta=\sin^{-1}\left(\frac35\right). $$

Then

$$ \sin\theta=\frac35, \qquad 0\le\theta\le\frac{\pi}{2}. $$

Using a right-angled triangle, we get

$$ \cos\theta=\frac45. $$

Therefore,

$$ \tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{3}{4}. $$

Using the double-angle formula,

$$ \tan2\theta = \frac{2\tan\theta}{1-\tan^2\theta}, $$

we obtain

$$ \tan2\theta = \frac{2\left(\frac34\right)} {1-\left(\frac34\right)^2} = \frac{\frac32} {\frac{7}{16}} = \frac{24}{7}. $$

Since

$$ 0\le2\theta\le\pi $$

and

$$ \tan2\theta>0, $$

it follows that

$$ 0<2\theta<\frac{\pi}{2}, $$

which lies in the principal value interval of

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

Therefore,

$$ 2\theta = \tan^{-1}\left(\frac{24}{7}\right). $$

Substituting

$$ \theta=\sin^{-1}\left(\frac35\right), $$

we get

$$ \boxed{2\sin^{-1}\left(\frac35\right) = \tan^{-1}\left(\frac{24}{7}\right).} $$

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Question 4

Prove that

$$ \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac35\right) = \tan^{-1}\left(\frac{77}{36}\right). $$

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Proof

Let

$$ A=\sin^{-1}\left(\frac{8}{17}\right), \qquad B=\sin^{-1}\left(\frac35\right). $$

Then

$$ \sin A=\frac{8}{17}, \qquad \sin B=\frac35. $$

Using right-angled triangles, we obtain

$$ \cos A=\frac{15}{17}, \qquad \cos B=\frac45. $$

Hence,

$$ \tan A=\frac{8}{15}, \qquad \tan B=\frac34. $$

Using the addition formula,

$$ \tan(A+B) = \frac{\tan A+\tan B} {1-\tan A\tan B}, $$

we get

$$ \tan(A+B) = \frac{\frac{8}{15}+\frac34} {1-\frac{8}{15}\cdot\frac34}. $$

Simplifying,

$$ \tan(A+B) = \frac{\frac{77}{60}} {\frac35} = \frac{77}{36}. $$

Since

$$ A,B\in\left[0,\frac{\pi}{2}\right], $$

we have

$$ 0which is the principal value interval of

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

Therefore,

$$ A+B = \tan^{-1}\left(\frac{77}{36}\right). $$

Substituting the values of

$$ A \quad\text{and}\quad B, $$

we obtain

$$ \boxed{\sin^{-1}\left(\frac{8}{17}\right) +\sin^{-1}\left(\frac35\right) = \tan^{-1}\left(\frac{77}{36}\right).} $$

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Question 5

Prove that

$$ \cos^{-1}\left(\frac45\right) + \cos^{-1}\left(\frac{12}{13}\right) = \cos^{-1}\left(\frac{33}{65}\right). $$

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Proof

Let

$$ A=\cos^{-1}\left(\frac45\right), \qquad B=\cos^{-1}\left(\frac{12}{13}\right). $$

Then

$$ \cos A=\frac45, \qquad \cos B=\frac{12}{13}. $$

Since

$$ A,B\in[0,\pi], $$

and both cosine values are positive, we have

$$ 0Using right-angled triangles,

$$ \sin A=\frac35, \qquad \sin B=\frac5{13}. $$

Using the cosine addition formula,

$$ \cos(A+B)=\cos A\cos B-\sin A\sin B, $$

we get

$$ \cos(A+B) = \frac45\cdot\frac{12}{13} - \frac35\cdot\frac5{13}. $$

Simplifying,

$$ \cos(A+B) = \frac{48-15}{65} = \frac{33}{65}. $$

Also,

$$ 0which is the principal value interval of

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

Therefore,

$$ A+B = \cos^{-1}\left(\frac{33}{65}\right). $$

Hence,

$$ \boxed{\cos^{-1}\left(\frac45\right) +\cos^{-1}\left(\frac{12}{13}\right) = \cos^{-1}\left(\frac{33}{65}\right).} $$

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Question 6

Prove that

$$ \cos^{-1}\left(\frac{12}{13}\right) + \sin^{-1}\left(\frac35\right) = \sin^{-1}\left(\frac{56}{65}\right). $$

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Proof

Let

$$ A=\cos^{-1}\left(\frac{12}{13}\right), \qquad B=\sin^{-1}\left(\frac35\right). $$

Then

$$ \cos A=\frac{12}{13}, \qquad \sin B=\frac35. $$

Since

$$ A,B\in\left[0,\frac{\pi}{2}\right], $$

we obtain

$$ \sin A=\frac5{13}, \qquad \cos B=\frac45. $$

Using the sine addition formula,

$$ \sin(A+B) = \sin A\cos B+\cos A\sin B, $$

we get

$$ \sin(A+B) = \frac5{13}\cdot\frac45 + \frac{12}{13}\cdot\frac35. $$

Simplifying,

$$ \sin(A+B) = \frac{20+36}{65} = \frac{56}{65}. $$

Also,

$$ 0which lies in the principal value interval of

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

Therefore,

$$ A+B = \sin^{-1}\left(\frac{56}{65}\right). $$

Hence,

$$ \boxed{\cos^{-1}\left(\frac{12}{13}\right) +\sin^{-1}\left(\frac35\right) = \sin^{-1}\left(\frac{56}{65}\right).} $$

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Question 7

Prove that

$$ \tan^{-1}\left(\frac{63}{16}\right) = \sin^{-1}\left(\frac5{13}\right) + \cos^{-1}\left(\frac35\right). $$

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Proof

Let

$$ A=\sin^{-1}\left(\frac5{13}\right), \qquad B=\cos^{-1}\left(\frac35\right). $$

Then

$$ \sin A=\frac5{13}, \qquad \cos B=\frac35. $$

Using right-angled triangles, we obtain

$$ \cos A=\frac{12}{13}, \qquad \sin B=\frac45. $$

Hence,

$$ \tan A=\frac5{12}, \qquad \tan B=\frac43. $$

Using the tangent addition formula,

$$ \tan(A+B) = \frac{\tan A+\tan B} {1-\tan A\tan B}, $$

we get

$$ \tan(A+B) = \frac{\frac5{12}+\frac43} {1-\frac5{12}\cdot\frac43}. $$

Simplifying,

$$ \tan(A+B) = \frac{\frac74}{\frac49} = \frac{63}{16}. $$

Also,

$$ 0which lies in the principal value interval of

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

Therefore,

$$ A+B = \tan^{-1}\left(\frac{63}{16}\right). $$

Hence,

$$ \boxed{\tan^{-1}\left(\frac{63}{16}\right) = \sin^{-1}\left(\frac5{13}\right) + \cos^{-1}\left(\frac35\right).} $$

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Question 8

Prove that

$$ \tan^{-1}\sqrt{x} = \frac12\cos^{-1}\left(\frac{1-x}{1+x}\right), \qquad x\in[0,1]. $$

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Proof

Let

$$ \theta=\tan^{-1}\sqrt{x}. $$

Then

$$ \tan\theta=\sqrt{x}, \qquad 0\le\theta\le\frac{\pi}{4}, $$

since

$$ 0\le x\le1. $$

Using the identity

$$ \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}, $$

and substituting

$$ \tan\theta=\sqrt{x}, $$

we get

$$ \cos2\theta = \frac{1-x}{1+x}. $$

Since

$$ 0\le\theta\le\frac{\pi}{4}, $$

it follows that

$$ 0\le2\theta\le\frac{\pi}{2}, $$

which lies in the principal value interval of

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

Therefore, applying

$$ \cos^{-1} $$

to both sides gives

$$ 2\theta = \cos^{-1}\left(\frac{1-x}{1+x}\right). $$

Dividing both sides by

$$ 2, $$

we obtain

$$ \theta = \frac12\cos^{-1}\left(\frac{1-x}{1+x}\right). $$

Since

$$ \theta=\tan^{-1}\sqrt{x}, $$

we get

$$ \boxed{\tan^{-1}\sqrt{x} = \frac12\cos^{-1}\left(\frac{1-x}{1+x}\right).} $$

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Question 9

Prove that

$$ \cot^{-1}\left( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}} {\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right) = \frac{x}{2}, \qquad x\in\left(0,\frac{\pi}{4}\right). $$

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Proof

Let

$$ A= \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}} {\sqrt{1+\sin x}-\sqrt{1-\sin x}}. $$

Using the half-angle identities,

$$ \sqrt{1+\sin x} = \sin\frac{x}{2} + \cos\frac{x}{2}, $$

and

$$ \sqrt{1-\sin x} = \cos\frac{x}{2} - \sin\frac{x}{2}, $$

because

$$ 0which implies

$$ 0<\frac{x}{2}<\frac{\pi}{8}, $$

and hence both

$$ \sin\frac{x}{2} \quad\text{and}\quad \cos\frac{x}{2} $$

are positive.

Substituting these values in

$$ A, $$

we obtain

$$ A = \frac{\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right) +\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)} {\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right) -\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}. $$

Simplifying,

$$ A = \frac{2\cos\frac{x}{2}} {2\sin\frac{x}{2}} = \cot\frac{x}{2}. $$

Therefore, the given expression becomes

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

Since

$$ 0<\frac{x}{2}<\frac{\pi}{8}<\pi, $$

the angle

$$ \frac{x}{2} $$

lies in the principal value interval of

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

namely

$$ (0,\pi). $$

Hence,

$$ \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}. $$

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Hence proved.

$$ \boxed{ \cot^{-1}\left( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}} {\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right) = \frac{x}{2} } $$

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Question 10

Prove that

$$ \tan^{-1}\left( \frac{\sqrt{1+x}-\sqrt{1-x}} {\sqrt{1+x}+\sqrt{1-x}} \right) = \frac{\pi}{4} - \frac12\cos^{-1}x, $$

where

$$ -\frac{1}{\sqrt2}\le x\le1. $$

Hint: Put

$$ x=\cos2\theta. $$

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Proof

Let

$$ x=\cos2\theta. $$

Since

$$ -\frac{1}{\sqrt2}\le x\le1, $$

we have

$$ 0\le2\theta\le\frac{3\pi}{4}, $$

or

$$ 0\le\theta\le\frac{3\pi}{8}. $$

Hence,

$$ \sin\theta\ge0 \quad\text{and}\quad \cos\theta\ge0. $$

Using the half-angle identities,

$$ 1+\cos2\theta=2\cos^2\theta, $$

and

$$ 1-\cos2\theta=2\sin^2\theta. $$

Therefore,

$$ \sqrt{1+x} = \sqrt{2}\cos\theta, $$

and

$$ \sqrt{1-x} = \sqrt{2}\sin\theta. $$

Substituting these values into the given expression,

$$ \begin{aligned} \tan^{-1}\left( \frac{\sqrt{1+x}-\sqrt{1-x}} {\sqrt{1+x}+\sqrt{1-x}} \right) &= \tan^{-1}\left( \frac{\sqrt2\cos\theta-\sqrt2\sin\theta} {\sqrt2\cos\theta+\sqrt2\sin\theta} \right)\\[2mm] &= \tan^{-1}\left( \frac{\cos\theta-\sin\theta} {\cos\theta+\sin\theta} \right). \end{aligned} $$

Using the identity

$$ \tan(A-B) = \frac{\tan A-\tan B} {1+\tan A\tan B}, $$

with

$$ A=\frac{\pi}{4}, \qquad B=\theta, $$

we obtain

$$ \frac{\cos\theta-\sin\theta} {\cos\theta+\sin\theta} = \tan\left(\frac{\pi}{4}-\theta\right). $$

Hence,

$$ \tan^{-1}\left( \frac{\cos\theta-\sin\theta} {\cos\theta+\sin\theta} \right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4}-\theta\right)\right). $$

Since

$$ 0\le\theta\le\frac{3\pi}{8}, $$

we have

$$ -\frac{\pi}{8} \le \frac{\pi}{4}-\theta \le \frac{\pi}{4}, $$

which lies completely inside the principal value interval of

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

namely

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

Therefore,

$$ \tan^{-1}\left(\tan\left(\frac{\pi}{4}-\theta\right)\right) = \frac{\pi}{4}-\theta. $$

Also, from

$$ x=\cos2\theta, $$

we get

$$ 2\theta=\cos^{-1}x, $$

because

$$ 0\le2\theta\le\pi, $$

which is the principal value interval of

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

Hence,

$$ \theta=\frac12\cos^{-1}x. $$

Substituting this value of

$$ \theta, $$

we obtain

$$ \boxed{ \tan^{-1}\left( \frac{\sqrt{1+x}-\sqrt{1-x}} {\sqrt{1+x}+\sqrt{1-x}} \right) = \frac{\pi}{4} - \frac12\cos^{-1}x. } $$

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Question 11

Solve the equation:

$$ 2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x). $$

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Solution

Let

$$ \theta=\tan^{-1}(\cos x). $$

Then

$$ \tan\theta=\cos x. $$

Using the double-angle formula,

$$ \tan2\theta = \frac{2\tan\theta}{1-\tan^2\theta}, $$

we get

$$ \tan\left(2\tan^{-1}(\cos x)\right) = \frac{2\cos x}{1-\cos^2x}. $$

Since

$$ 1-\cos^2x=\sin^2x, $$

the above expression becomes

$$ \frac{2\cos x}{\sin^2x}. $$

Taking tangent on both sides of the given equation,

$$ \frac{2\cos x}{\sin^2x} = 2\cosec x = \frac{2}{\sin x}. $$

Therefore,

$$ \frac{\cos x}{\sin^2x} = \frac{1}{\sin x}. $$

Multiplying both sides by

$$ \sin^2x, $$

we obtain

$$ \cos x=\sin x. $$

Hence,

$$ \tan x=1. $$

Therefore, the general solution is

$$ \boxed{x=n\pi+\frac{\pi}{4},\qquad n\in\mathbb{Z}.} $$

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Question 12

Solve the equation:

$$ \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac12\tan^{-1}x, \qquad x>0. $$

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Solution

Let

$$ \theta=\frac12\tan^{-1}x. $$

Then

$$ 2\theta=\tan^{-1}x, $$

which gives

$$ \tan2\theta=x. $$

Using the double-angle formula,

$$ \tan2\theta = \frac{2\tan\theta}{1-\tan^2\theta}, $$

we obtain

$$ x=\frac{2t}{1-t^2}, $$

where

$$ t=\tan\theta. $$

Also, from the given equation,

$$ \tan^{-1}\left(\frac{1-x}{1+x}\right)=\theta, $$

so that

$$ t=\frac{1-x}{1+x}. $$

Substituting this value of

$$ t $$

into

$$ x=\frac{2t}{1-t^2}, $$

we get

$$ x = \frac{2\left(\frac{1-x}{1+x}\right)} {1-\left(\frac{1-x}{1+x}\right)^2}. $$

Now,

$$ 1-\left(\frac{1-x}{1+x}\right)^2 = \frac{(1+x)^2-(1-x)^2}{(1+x)^2} = \frac{4x}{(1+x)^2}. $$

Hence,

$$ x = \frac{2(1-x)}{1+x}\cdot\frac{(1+x)^2}{4x} = \frac{(1-x)(1+x)}{2x} = \frac{1-x^2}{2x}. $$

Therefore,

$$ 2x^2=1-x^2, $$

which gives

$$ 3x^2=1. $$

Thus,

$$ x=\pm\frac1{\sqrt3}. $$

Since

$$ x>0, $$

the required solution is

$$ \boxed{x=\frac1{\sqrt3}}. $$

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Question 13

Find the value of

$$ \sin\left(\tan^{-1}x\right), \qquad |x|<1. $$

Choose the correct answer.

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Solution

Let

$$ \theta=\tan^{-1}x. $$

Then

$$ \tan\theta=x=\frac{\text{Perpendicular}}{\text{Base}}. $$

Take a right triangle with perpendicular

$$ x, $$

and base

$$ 1. $$

By the Pythagoras theorem, the hypotenuse is

$$ \sqrt{1+x^2}. $$

Therefore,

$$ \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{x}{\sqrt{1+x^2}}. $$

Since

$$ \theta=\tan^{-1}x, $$

we obtain

$$ \boxed{\sin\left(\tan^{-1}x\right)=\frac{x}{\sqrt{1+x^2}}.} $$

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Correct Option:

$$ \boxed{\text{(D)}} $$

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Question 14

If

$$ \sin^{-1}(1-x)-2\sin^{-1}x=\frac{\pi}{2}, $$

find the value of

$$ x. $$

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Solution

Let

$$ \theta=\sin^{-1}x. $$

Then

$$ \sin^{-1}(1-x) = \frac{\pi}{2}+2\theta. $$

Since the principal value of

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

always lies in the interval

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

the value

$$ \frac{\pi}{2}+2\theta $$

must also lie in this interval.

This is possible only when

$$ \theta=0. $$

Hence,

$$ x=\sin0=0. $$

Substituting

$$ x=0 $$

into the given equation,

$$ \sin^{-1}(1)-2\sin^{-1}(0) = \frac{\pi}{2}-0 = \frac{\pi}{2}, $$

which satisfies the equation.

Now check

$$ x=\frac12. $$

Then

$$ \sin^{-1}\left(\frac12\right) - 2\sin^{-1}\left(\frac12\right) = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \neq \frac{\pi}{2}. $$

Hence,

$$ x=\frac12 $$

is not a solution.

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Answer:

$$ \boxed{x=0.} $$

Correct Option:

$$ \boxed{\text{(C)}} $$

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