# 25. Inverse Circular Functions¶

Definition: Inverse functions related to trigonometrical ratios are called inverse trigonometric functions. The definition of different inverse trigonometric functions is given below:

If $\sin\theta = x,$ then $\theta = \sin^{-1}x,$ provided $-1\leq x\leq 1$ and $-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}$

If $\cos\theta = x,$ then $\theta = \cos^{-1}x,$ provoded $-1\leq x\leq 1$ and $0\leq \theta \leq \pi$

If $\tan\theta = x,$ then $\theta = \tan^{-1}x,$ provided $-\infty < x < \infty$ and $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$

If $\cot\theta = x,$ then $\theta = \cot^{-1}x,$ provided $-\infty < x < \infty$ and $0 < \theta < \pi$

If $\sec\theta = x,$ then $\theta = \sec^{-1}x,$ provided $x \leq -1$ or $x\geq 1$ and $0\leq \theta \leq \pi, \theta \neq \frac{\pi}{2}$

If $\cosec\theta = x,$ then $\theta = \cosec^{-1}x,$ provided $x \leq -1$ or $x\geq 1$ and $x\leq -1$ or $x\geq 1$ and $-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2}, \theta\neq 0$

Note: In the above definition restrictions on $\theta$ are due to the consideration of principal values of inverse terms. If these restrictions are removed, the terms will represent inverse trigonometrical relation and not functions.

Notations: I. Arc $\sin x$ denotes the sine inverse of $x$ [General value]

$\arcsin x$ denotes the sine inverse of $x$ [Principal value]

II. $\sin^{-1}x$ denotes the principal value of sine inverse $x$

From the above notations three imprtant results follow:

1. $\sin^{-1}x = \theta \Rightarrow \sin \theta = x$ and $\theta$ is the principal value.

2. $\sin^{-1}x = \arcsin x, \cos^{-1}x = \arccos x$

3. From the definition of the inverse functions, we know that if $y = f(x)$ is a function then for $f^{-1}$ to be a function, $f$ must be one-one and onto mapping.

When we consider $y = Arc\sin x,$ for any $x\in[-1, 1]$ infinite number of values of $y$ are obtained and hence it does not represent inverse functions. When $y = \arcsin x$ or $\sin^{-1}x,$ corresponding to one value of $x\in [-1, 1],$ one values of $y$ is obtained and hence it represents inverse trigonometric function.

Hence, for inverse trigonometric functions, consideration of principal values is essential.

## 25.1. Principal Value¶

Numerically smallest angle is known as the principal value.

Since inverse trigonometrical terms are in fact angles, definitions of principal value of inverse trigonometrical term is the same as the definition of the principal value of angles.

Suppose we have to find the principal value of $\sin^{-1}\frac{1}{2}.$

For this let $\sin^{-1}\frac{1}{2} = \theta$ then $\sin\theta = \frac{1}{2}$

$\Rightarrow \theta = \ldots, -\frac{11\pi}{6}, -\frac{7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \ldots$

Among all these angles $\frac{\pi}{6}$ is the numerically smallest angles satisfying $\sin\theta = \frac{1}{2}$ and hence it is principal value.

The steps to find principal value is same as described in previous chapter.

## 25.2. Important Formulae¶

1. $\sin\sin^{-1}x = x, -1\leq x\leq 1$

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

3. $\tan\tan^{-1}x = x, -\infty< x < \infty$

4. $\cot\cot^{-1}x = x, -\infty< x < \infty$

5. $\sec\sec^{-1}x = x, x\leq -1$ or $x\geq 1$

6. $\cosec\cosec^{-1}x = x, x\leq -1$ or $x\geq 1$

Proof: Let $\sin^{-1}x = \theta$ then $\sin\theta = x$

Putthing the value of $\theta$ from first equation in second

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

Other formulae can be proved in similar manner.

1. $\sin^{-1}\sin x = x~\forall~-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$

2. $\cos^{-1}\cos x = x~\forall~0\leq x\leq \pi$

3. $\tan^{-1}\tan x = x~\forall~0

4. $\cot^{-1}\cot x = x~\forall~0\leq x\leq \pi$

5. $\sec^{-1}\sec x = x~\forall~0\leq x\leq\pi, x\neq\frac{\pi}{2}$

6. $\cosec^{-1}\cosec x = x~\forall~-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}, x\neq 0$

Proof: Let $\sin^{-1}x = \theta$ then $\sin\theta = x$

Substituting the value of $x,$ we get

$\sin^{1}\sin\theta = \theta$

Replacing $\theta$ by $x,$ we get

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

Other formulae can be proved in similar manner.

1. $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}~\forall~-1\leq x\leq 1$

2. $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}~\forall x \in R$

3. $\sec^{-1}x + \cosec^{-1}x = \frac{\pi}{2}~\forall x\leq -1$ or $x\geq 1$

Proof: Let $\sin^{-1}x = \theta$ then $\sin\theta = x$

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

$\cos^{-1}x + \theta = \frac{\pi}{2}\Rightarrow \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$

Similarly other results can be proven.

1. $\sin^{-1}x = \cosec^{-1}\frac{1}{x}, -1\leq x\leq 1$

$\cosec^{-1}x = \sin^{-1}\frac{1}{2}, x\leq -1$ or $x > 1$

2. $\cos^{-1}x = sec^{-1}\frac{1}{x}, -1\leq x\leq 1$

$\sec^{-1}x = \cos^{-1}\frac{1}{x}, x\leq -1$ or $x\geq 1$

3. $\tan^{-1}x = \cot^{-1}\frac{1}{x}, x>0$

$\cot^{-1}x = \tan^{-1}\frac{1}{x}, x >0$

$\tan^{-1}x = \cot^{-1}\frac{1}{x} - \pi, x < 0$

$\cot^{-1}x = \pi + \tan^{-1}\frac{1}{x}, x < 0$

Proof: Let $\sin^{-1}x = \theta$ then $\sin\theta = x$

$\Rightarrow \cosec\theta = \frac{1}{x}$

$\Rightarrow \theta = \cosec^{-1}\frac{1}{x}$

$\Rightarrow \sin^{-1}x = \cosec^{-1}\frac{1}{x}$

Other results can be proven similarly.

1. $\sin^{-1}x = \cos^{-1}\sqrt{1 - x^2}, ~\forall~0\leq x\leq 1$

2. $\sin^{-1}x = -\cos^{-1}\sqrt{1 - x^2}~forall~-1\leq x< 0$

Proof: Let $\sin^{-1}x = \theta$ then $\sin\theta = x$

$\Rightarrow \cos^2\theta = 1 - x^2 \Rightarrow \cos\theta = \pm\sqrt{1 - x^2}$

Principal values of $\sin^{-1}x$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$

In this interval $\cos\theta$ is +ve

$\Rightarrow \sin^{-1}x = \cos^{-1}\sqrt{1 - x^2}$

For $-1\leq x < 0$ $\sin^{-1}x$ will be negative angle while $\cos^{-1}\sqrt{1 - x^2}$ will be positive angle. Hence to balance that we need to used a negative sign for this.

1. $\sin^{-1}(-x) = -\sin^{-1}x$

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

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

4. $\cot^{-1}x = \pi - \cot^{-1}x$

Proof: Let $\cos^{-1}(-x) = \theta$ then $\cos\theta = -x$

$-\cos\theta = x \Rightarrow \cos(\pi - \theta) = x$

$\therefore \theta = \pi - \cos^{-1}x$

Note: $\cos(\pi + \theta)$ is also equal to $-\cos\theta$ but this will make principal value greater than $\pi.$

Similarly other results can be proven.

1. $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy}$ where $x, y > 0$ and $xy < 1$

$\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x + y}{1 - xy}$ where $x, y > 0$ and $xy > 1$

$\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x + y}{1 - xy}$ where $x, y , 0$ and $xy > 1$

2. $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x - y}{1 + xy}$ where $xy > 1$

Proof: Let $\tan^{-1}x = \alpha$ and $\tan^{-1}y = \beta$ then

$\tan\alpha = x$ and $\tan\beta = y$

$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{x + y}{1 - xy}$

$\Rightarrow \alpha +\beta = \tan^{-1}\frac{x + y}{1 = xy}$

$\Rightarrow \tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy}$

Case I. When $x, y > 0$ and $xy < 1, \tan^{-1}\frac{x + y}{1 - xy} > 0$

therefore $\tan^{-1}\frac{x + y}{1 - xy}$ will be a positive angle.

Case II. When $x, y >0$ and $xy > 1$ $\tan^{-1}\frac{x + y}{1 - xy}$ will be a negative angle.

$\therefore \tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x + y}{1 - xy}$

Case III. When $x, y< 0$ and $xy > 1$ $\tan^{-1}x + \tan^{-1}y$ will be a negative angle and $\tan^{-1}\frac{x + y}{1 - xy}$ will be a positive angle.

To balance it we will need to add $-\pi$

$\therefore \tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x + y}{1 - xy}$

Similarly other result can be proven.

1. $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \frac{x + y + z - xyz}{1 - xy - yz - xz}$

This can be proven that previous formula.

1. $\sin^{-1}x + \sin^{-1}y = \sin^{-1}[x\sqrt{1 - y^2} + y\sqrt{1 - x^2}]$ if $-1\leq x, y\leq 1$ and $x^2 + y^2\leq 1$ or if $xy < 0$ and $x^2 + y^2 > 1$

2. $\sin^{-1}x - \sin^{-1}y = \sin^{-1}[x\sqrt{1 - y^2} - y\sqrt{1 - x^2}]$ if $-1\leq x, y\leq 1$ and $x^2 + y^2\leq 1$ or if $xy > 0$ and $x^2 + y^2 > 1$

Proof: Let $\sin^{-1}x = \alpha$ and $\sin^{-1}y = \beta$ then $\sin\alpha = x, \sin\beta = y.$

Now $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \sin\beta\cos\alpha$

$= \sin\alpha\sqrt{1 - \sin^2\beta} + \sin\beta\sqrt{1 - \sin^2\alpha}$

$= x\sqrt{1 - y^2} + y\sqrt{1 - x^2}$

$\alpha + \beta = \sin^{-1}[x\sqrt{1 - y^2} + y\sqrt{1 - x^2}]$

Similarly we can prove that $\sin^{-1}x - \sin^{-1}y = \sin^{-1}[x\sqrt{1 - y^2} - y\sqrt{1 - x^2}]$

1. $2\tan^{-1}x = \sin^{-1}\frac{2x}{1 + x^2},$ where $|x|< 1$

2. $2\tan^{-1}x = \cos^{-1}\frac{1 - x^2}{1 + x^2},$ where $x\geq 0$

3. $2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^2},$ where $|x| < 1$

Proof:

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

$\sin2\theta = \frac{2\tan\theta}{1 + \tan^2x\theta} = \frac{2x}{1 + x^2}$

$\Rightarrow 2\theta = \sin^{-1}\frac{2x}{1 + x^2}\Rightarrow 2\tan^{-1}x = \sin^{-1}\frac{2x}{1 + x^2}$

Here, $-\frac{\pi}{2}\leq \sin^{-1} \leq \frac{\pi}{2}$

$-\frac{\pi}{2}\leq 2\tan^{-1}x\leq \frac{\pi}{2}$

$-\frac{\pi}{4}\leq \tan^{-1}x\leq \frac{\pi}{4}$

$-1\leq x\leq 1 \Rightarrow |x| < 1$

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

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

$2\tan^{-1}x = \cos^{-1}\frac{1 - x^2}{1 + x^2}$

For $x \geq 0$ both sides will be balanced.

For $x<0, 2\tan^{-1}x$ will represent a negative angle where R.H.S. will always lie between $0$ and $\pi.$ Hence two sides cannot be equal.

3. $\tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2x}{1 - x^2}\Rightarrow 2\theta = \tan^{-1}\frac{2x}{1 - x^2}$

$2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^2}$ which holds good for $|x|< 1$

1. $2\sin^{-1}x = \sin^{-1}[2x\sqrt{1 - x^2}]$ if $-\frac{1}{\sqrt{2}}\leq x\leq \frac{1}{\sqrt{2}}$

2. $2\cos^{-1}x = \cos^{-1}(2x^2 - 1)$ where $0\leq x \leq 1$

These can be proven like $\sin^{-1}x + \sin^{-1}y$

1. $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$ where $-\frac{1}{2}\leq x\leq \frac{1}{2}$

2. $3\cos^{-1}x = \cos^{-1}(4x^3 - 3x)$ where $\frac{1}{2}\leq x\leq 1$

3. $3\tan^{-1}x = \tan^{-1}\frac{3x - x^3}{1 - 3x^2}$ where $-\frac{1}{\sqrt{3}}< x < \frac{1}{\sqrt{3}}$

These can be proven like previous proof.

## 25.3. Graph of Important Inverse Trigonometric Functions¶

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

Frpm this graph we observer following:

1. Domain is $-1\leq x\leq 1$

2. Range is $-\frac{\pi}{2}\leq y \leq \frac{\pi}{2}$

3. $\because \sin^{-1}x = -\sin^{-1}x \therefore y = \sin^{-1}x$ is an odd function.

4. It is a non-periodic function

5. It passes through origin i.e. when $x = 0, y = 0$

2. $y = \cos^{-1}x, -1\leq x\leq 1$

Follwing points can be observed from the graph:

1. Domain is $-1\leq x\leq 1$

2. Range is $0\leq x\leq \pi$

3. $\because \cos^{-1}(-x) = \pi - \cos^{-1}x$

$\Rightarrow y = \cos^{-1}x$ is neither odd nor even.

4. It is a non-periodic function

3. $y = \tan^{-1}x, -\infty< x <\infty$

From the graph follwing points can be observed:

1. Domain is $-\infty < x < \infty$

2. Range is $-\frac{\pi}{2} < x < \frac{\pi}{2}$

3. $y = \tan^{-1}x$ is an odd function

4. It is a non-periodic function.

5. It passes through origin.

4. $y = \cot^{-1}x, -\infty< x <\infty$

From the graph follwing points can be observed:

1. Domain is $-\infty < x < \infty$

2. Rnage is $0

3. The function is neither odd nor even.

4. It is a non-periodic function

## 25.4. Problems¶

Evaluate the following:

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

2. $\cot^{-1}(-1)$

3. $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

Find the value of the following:

1. $\sin\left[\frac{\pi}{3} - \sin^{-1}\frac{-1}{2}\right]$

2. $\sin\left[\cos^{-1}\frac{-1}{2}\right]$

3. $\sin\left[\tan^{-1}(-\sqrt{3}) + \cos^{-1}\frac{-\sqrt{3}}{2}\right]$

4. Evaluate $\tan\left[\frac{1}{2}\cos^{-1}\frac{\sqrt{5}}{3}\right]$

5. Find the angle $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$

Find the value of the following:

1. $\sin^{-1}\frac{\sqrt{3}}{2}$

2. $\tan^{-1}\frac{-1}{\sqrt{3}}$

3. $\cot^{-1}(-\sqrt{3})$

4. $\cot^{-1}\cot\frac{5\pi}{4}$

5. $\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$

6. $\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2}$

7. $\cos\left[\tan^{-1}\left(\frac{3}{4}\right)\right]$

8. $\cos\left[\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{6}\right]$

9. Prove that $2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4}$

10. Prove that $\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}$

11. Prove that $\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65} = \frac{\pi}{2}$

12. Prove that $4\tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{70} + \tan^{-1}\frac{1}{99} = \frac{\pi}{4}$

13. Prove that $\cot^{-1}9 + \cosec^{-1}\frac{\sqrt{41}}{4} = \frac{\pi}{4}$

14. Prove that $4(\cot^{-1}3 + \cosec^{-1}\sqrt{5}) = \pi$

15. Prove that $\tan^{-1}x = 2\tan^{-1}[\cosec\tan^{-1}x - \tan\cot^{-1}x]$

16. Prove that $2\tan^{-1}\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \cos^{-1}\left[\frac{b + a\cos x}{a + b\cos x}\right]$ for $0 and $x\geq 0.$

17. Prove that $\tan^{-1}\frac{x - y}{1 + xy} + \tan^{-1}\frac{y - z}{1 + yz} + \tan^{-1}\frac{z - x}{1 + zx} = \tan^{-1}\left(\frac{x^2 - y^2}{1 + x^2y^2}\right) + \tan^{-1}\left(\frac{y^2 - z^2}{1 + y^2z^2}\right) + \tan^{-1}\left(\frac{z^2 - x^2}{1 + z^2x^2}\right)$

18. Prove that $\sin\cot^{-1}\tan\cos^{-1}x = x$

19. Prove that $\tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}(\cot x) +\tan^{-1}(\cot^3x) = 0$ if $\frac{\pi}{4}< x < \frac{\pi}{2}, = \pi$ if $0

20. Prove that $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{5} + \tan^{-1}\frac{1}{4} = \frac{\pi}{4}$

21. Prove that $\tan^{-1}\frac{2a - b}{\sqrt{3}b} + \tan^{-1}\frac{2b - a}{\sqrt{3}a} = \frac{\pi}{3}$

22. Prove that $\tan^{-1}\frac{2}{5} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{12} = \frac{\pi}{4}$

23. Prove that $2\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{32}{43}$

24. Prove that $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi = 2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)$

25. Prove that $\tan^{-1}x + \cot^{-1}y = \tan^{-1}\frac{xy + 1}{y - x}$

26. Prove that $\tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1} = \cot^{-1}x$

27. Prove that $2\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \pi/4$

28. Prove that $\tan^{-1}\frac{a - b}{1 + ab} + \tan^{-1}\frac{b - c}{1 + bc} + \tan^{-1}\frac{c - a}{1 + ca} = 0$

29. Prove that $\tan^{-1}\frac{a^3 - b^3}{1 + a^3b^3} + \tan^{-1}\frac{b^3 - c^3}{1 + b^3c^3} + \tan^{-1}\frac{c^3 - a^3}{1 + c^3a^3} = 0$

30. Prove that $\cot^{-1}\frac{xy + 1}{y - x} + \cot^{-1}\frac{yz + 1}{z - y} + \cot^{-1}z = \tan^{-1}\frac{1}{x}$

31. Prove that $\cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right) = 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)$

32. Prove that $\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}$

33. Prove that $\cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65} = \frac{\pi}{2}$

34. Prove that $\sin^{-1}x + \sin^{-1}y = \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right)$ where $x, y \in[0, 1]$

35. Prove that $4\left(\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}}\right) =\pi$

36. Prove that $\cos(2\sin^{-1}x) = 1 - 2x^2$

37. Prove that $\frac{1}{2}\cos^{-1}x = \sin^{-1}\sqrt{\frac{1 - x}{2}} = \cos^{-1}\sqrt{\frac{1 + x}{2}} = \tan^{-1}\frac{\sqrt{1 - x^2}}{1 + x}$

38. Prove that $\sin^{-1}x + \cos^{-1}y = \tan^{-1}\frac{xy + \sqrt{(1 - x^2)(1 - y^2)}}{y\sqrt{1 - x^2} - x\sqrt{1 - y^2}}$

39. Prove that $\tan^{-1}x + \tan^{-1}y = \frac{1}{2}\sin^{-1}\frac{2(x + y)(1 - xy)}{(1 + x^2)(1 + y^2)}$

40. Prove that $2\tan^{-1}(\cosec\tan^{-1}x - \tan\cot^{-1}x) = \tan^{-1}x$

41. Prove that $\cos\tan^{-1}\sin\cot^{-1}x = \sqrt{\frac{x^2 + 1}{x^2 + 2}}$

42. In any $\triangle ABC$ if $A = \tan^{-1}2$ and $B = \tan^{-1}3,$ prove that $C = \frac{\pi}{4}$

43. If $\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi$ then prove that $x^2 + y^2 + z^2 + 2xyz = 1$

44. If $\cos^{-1}\frac{x}{2}+ \cos^{-1}\frac{y}{3} = \theta,$ prove that $9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta$

45. If $r = x + y + z$ then prove that $\tan^{-1}\sqrt{\frac{xr}{yz}} + \tan^{-1}\sqrt{\frac{yr}{xz}} + \tan^{-1}\sqrt{\frac{zr}{xy}} = \pi$

46. If $u = \cot^{-1}\sqrt{\cos2\theta} - \tan^{-1}\sqrt{\cos2\theta}$ then prove that $\sin u = \tan^2\theta$

47. Solve $\cos^{-1}x\sqrt{3} + \cos^{-1}x = \frac{\pi}{2}$

48. Solve $\sin^{-1}x + \sin^{-1}2x = \frac{\pi}{3}$

49. If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z= \frac{\pi}{2},$ prove that $xy + yz + zx = 1$

50. If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z= \pi,$ prove that $x + y + z = xyz$

51. If $\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2},$ prove that $x\sqrt{1 - y^2} + y\sqrt{1 - x^2} = 1$

52. If $\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi,$ prove that $x\sqrt{1 - x^2} + y\sqrt{1 - y^2} + z\sqrt{1 - z^2} = 2xyz$

53. Establish the relationship between $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are in A.P. and if further $x, y, z$ are also in A.P. then prove that $x = y = z.$

54. Solve for $x, \cot^{-1}x + \sin^{-1}\frac{1}{\sqrt{5}} = \frac{\pi}{4}$

55. Solve $\tan^{-1}2x + \tan^{-1}3x = \frac{\pi}{4}$

56. Solve $\tan^{-1} x + \tan^{-1}\frac{2x}{1 - x^2} = \frac{\pi}{3}$

57. Solve $\tan^{-1}\frac{1}{2} = \cot^{-1}x + \tan^{-1}\frac{1}{7}$

58. Solve $\tan^{-1}(x - 1) + \tan^{-1}x + \tan^{-1}(x + 1) = \tan^{-1}3x$

59. Solve $\tan^{-1}\frac{x + 1}{x - 1} + \tan^{-1}\frac{x - 1}{x} = \pi + \tan^{-1}(-7)$

60. Solve $\cot^{-1}(a - 1) = \cot^{-1}x + \cot^{-1}(a^2 - x + 1)$

61. Solve $\sin^{-1}\frac{2\alpha}{1 + \alpha^2} + \sin^{-1}\frac{2\beta}{1 + \beta^2} = 2\tan^{-1}x$

62. Solve $\cos^{-1}\frac{x^2 - 1}{x^2 + 1} + \tan^{-1}\frac{2x}{x^2 - 1} = \frac{2\pi}{3}$

63. Solve $\sin^{-1}\frac{2a}{1 + a^2} + \cos^{-1}\frac{1 - b^2}{1 + b^2} = 2\tan^{-1}x$

64. Solve $\sin^{-1}x + \sin^{-1}(1 - x) = \cos^{-1}x$

65. Solve $\tan^{-1}ax + \frac{1}{2}\sec^{-1}bx = \frac{\pi}{4}$

66. Solve $\tan(\cos^{-1}x) = \sin(\tan^{-1}2)$

67. Solve $\tan\left(\sec^{-1}\frac{1}{x}\right) = \sin\cos^{-1}\frac{1}{\sqrt{5}}$

68. Find the values of $x$ and $y$ satisfying $\sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3}$ and $\cos^{-1}x - \cos^{-1}y = \frac{\pi}{3}$

69. Find the angle $\sin^{-1}(\sin10)$

70. Using principal values, express the following as a single angle $3\tan^{-1}\frac{1}{2} + 2\tan^{-1}\frac{1}{5} + \sin^{-1}\frac{142}{65\sqrt{5}}$

71. Find the value of $2\cos^{-1}x + \sin^{-1}x$ at $x = \frac{1}{5}$ where $0\leq \cos^{-1}x\leq \pi$ and $-\frac{\pi}{2}\leq \sin^{-1}x \leq \frac{\pi}{2}$.

72. Show that $\frac{1}{2}\cos^{-1}\frac{3}{5} = \tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\frac{4}{5}$

73. Find the greater angle between $2\tan^{-1}(2\sqrt{2} - 1)$ and $3\sin^{-1}\frac{1}{3} + \sin^{-1}\frac{3}{5}$

74. Prove that $\tan^{-1}\left(\frac{a_1x - y}{x + a_1y}\right) + \tan{-1}\left(\frac{a_2 - a_1}{1 + a_2a_1}\right) + \tan^{-1}\left(\frac{a3 - a_2}{1 + a_3a_2}\right) + \ldots + \tan^{-1}\left(\frac{a_n - a_{n - 1}}{1 + a_na_{n - 1}}\right) + \tan^{-1}\frac{1}{a_n} = \tan^{-1}\frac{x}{y}$

75. Find the sum $\cot^{-1}2 + \cot^{-1}8 + \cot^{-1}18 + \ldots +$ to $\infty$

76. Show that the function $y = 2\tan^{-1}x + \sin^{-1}\frac{2x}{1 + x^2}$ is constant for $x\geq 1.$ Find the value of this constant.

77. Prove the relations $\cos^{-1}x_0 = \frac{\sqrt{1 - x_0^2}}{x_1x_2x_3\ldots\text{~to~}\infty}$ where the successive quantities $x_r$ are connected by the relation $x_{r + 1} = \sqrt{\frac{1 + x_r}{2}}$ where $0\leq \cos^{-1}x_0\leq \pi$.

78. If $a, b$ are positive quantities and if $a_1 = \frac{a + b}{2}, b_1 = \sqrt{a_1b}, a_2 = \frac{a_1 + b_1}{2}, b_2 = \sqrt{a_2b_1}$ and so on then show that $\lim_{n\to \infty}a_n\lim_{n\to\infty}b_n = \frac{\sqrt{b^2 - a^2}}{\cos^{-1}\frac{a}{b}}$