Question 1

Find the principal values of the following:

1.  $\sin^{-1}\left(-\frac{1}{2}\right)$
2.  $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$
3.  $\csc^{-1}(2)$
4.  $\tan^{-1}(-\sqrt{3})$
5.  $\cos^{-1}\left(-\frac{1}{2}\right)$
6.  $\tan^{-1}(-1)$

7.  $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$
8.  $\cot^{-1}(\sqrt{3})$
9.  $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

10.  $\csc^{-1}(-\sqrt{2})$

Accepted Answer

1. $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ because principal value of $\sin^{-1}x$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

2. $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$ since principal value of $\cos^{-1}x$ lies in $[0, \pi]$.

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

4. $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$ as principal value of $\tan^{-1}x$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$.

5. $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.

6. $\tan^{-1}(-1) = -\frac{\pi}{4}$.

7. $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.

8. $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.

9. $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}$.

10. $\csc^{-1}(-\sqrt{2}) = \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$. — Use the principal value ranges:
- $\sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
- $\cos^{-1}x \in [0, \pi]$
- $\tan^{-1}x \in (-\frac{\pi}{2}, \frac{\pi}{2})$
- $\sec^{-1}x = \cos^{-1}\frac{1}{x}$
- $\csc^{-1}x = \sin^{-1}\frac{1}{x}$
- $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ with appropriate domain considerations.

Evaluate each inverse trig function using these ranges and known special angles.

Question 2

Find the values of the following:

11.  $\tan^{-1}(1) + \cos^{-1} - \frac{1}{2} + \sin^{-1} - \frac{1}{2}$
12.  $\cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2}$

Accepted Answer

11. Evaluate each term:
- $\tan^{-1}(1) = \frac{\pi}{4}$
- $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$
- $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$

Sum = $\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$.

12. $\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$, $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$

Therefore, $\cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2} = \frac{\pi}{3} + 2 \times \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$. — Use known values of inverse trig functions at special angles:
- $\tan^{-1}(1) = \frac{\pi}{4}$
- $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$
- $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$
- $\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$
- $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$

Add accordingly and simplify.

Question 3

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

(A)  $0 \leq y \leq \pi$

(B)  $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

(C)  $0 < y < \pi$

(D)  $-\frac{\pi}{2} < y < \frac{\pi}{2}$

Accepted Answer

$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

Explanation: The principal value branch of $\sin^{-1}x$ is defined such that $y = \sin^{-1}x$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. — By definition, the principal value of $\sin^{-1}x$ is the unique value $y$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin y = x$.

Question 4

$\tan^{-1}\sqrt{3} -\sec^{-1}(-2)$  is equal to

(A)  $\pi$

(B)  $-\frac{\pi}{3}$

(C)  $\frac{\pi}{3}$

(D)  $\frac{2\pi}{3}$

Accepted Answer

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

Explanation:
- $\tan^{-1}\sqrt{3} = \frac{\pi}{3}$
- $\sec^{-1}(-2) = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$

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

But since the options include $-\frac{\pi}{3}$ as (B), check carefully:
Actually, the question asks for the value, so the answer is (B) $-\frac{\pi}{3}$.

Hence correct answer is (B) $-\frac{\pi}{3}$. — Evaluate each inverse function:
- $\tan^{-1}\sqrt{3} = \frac{\pi}{3}$
- $\sec^{-1}(-2) = \cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$

Subtracting gives $\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$.

Therefore, option (B) is correct.

Question 5

Prove the following:

1. $3 \sin^{-1} x = \sin^{-1} (3x - 4x^3)$, $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$
2. $3 \cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, $x \in \left[\frac{1}{2}, 1\right]$

Accepted Answer

1. To prove: 3 sin^{-1} x = sin^{-1} (3x - 4x^3), for x in [-1/2, 1/2]

Let 	heta = \sin^{-1} x, so that x = \sin 	heta and 	heta \in [-\pi/6, \pi/6] because sin^{-1} x is defined in [-\pi/2, \pi/2] and x in [-1/2, 1/2].

Then, 3 \sin^{-1} x = 3	heta.

Using the triple angle formula for sine:
\sin 3	heta = 3 \sin 	heta - 4 \sin^3 	heta = 3x - 4x^3.

Therefore,

\sin (3 \sin^{-1} x) = 3x - 4x^3.

Since \sin^{-1} is the inverse sine function, and the range of 3	heta is within [-\pi/2, \pi/2] for the given x, we have:

3 \sin^{-1} x = \sin^{-1} (3x - 4x^3).

2. To prove: 3 \cos^{-1} x = \cos^{-1} (4x^3 - 3x), for x in [1/2, 1]

Let \phi = \cos^{-1} x, so that x = \cos \phi and \phi \in [0, \pi/3] because \cos^{-1} x is defined in [0, \pi] and x in [1/2, 1].

Then, 3 \cos^{-1} x = 3\phi.

Using the triple angle formula for cosine:
\cos 3\phi = 4 \cos^3 \phi - 3 \cos \phi = 4x^3 - 3x.

Since \cos^{-1} is the inverse cosine function, and the range of 3\phi is within [0, \pi], we have:

3 \cos^{-1} x = \cos^{-1} (4x^3 - 3x).

Hence, both identities are proved. — Used triple angle formulas for sine and cosine:
\sin 3	heta = 3 \sin 	heta - 4 \sin^3 	heta
\cos 3\phi = 4 \cos^3 \phi - 3 \cos \phi
Substituted 	heta = \sin^{-1} x and \phi = \cos^{-1} x, then used inverse functions' properties and domain restrictions to conclude the equalities.

Question 6

Write the following functions in the simplest form:

3. $	an^{-1}\frac{\sqrt{1 + x^2} - 1}{x}, x 
eq 0$
4. $	an^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}ight), 0 < x < \pi$
5. $	an^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}ight), \frac{-\pi}{4} < x < \frac{3\pi}{4}$
6. $	an^{-1}\frac{x}{\sqrt{a^2 - x^2}}, |x| < a$
7. $	an^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}ight), a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$

Accepted Answer

3. Simplify $	an^{-1}\frac{\sqrt{1 + x^2} - 1}{x}$, $x 
eq 0$:

Let 	heta = 	an^{-1} x \implies x = 	an 	heta.

Note that:
\sqrt{1 + x^2} = \sqrt{1 + 	an^2 	heta} = \sec 	heta.

So the expression inside the arctan is:
\frac{\sec 	heta - 1}{	an 	heta} = \frac{\sec 	heta - 1}{	an 	heta}.

Multiply numerator and denominator by \sec 	heta + 1:

= \frac{(\sec 	heta - 1)(\sec 	heta + 1)}{	an 	heta (\sec 	heta + 1)} = \frac{\sec^2 	heta - 1}{	an 	heta (\sec 	heta + 1)}.

Since \sec^2 	heta - 1 = 	an^2 	heta,

= \frac{	an^2 	heta}{	an 	heta (\sec 	heta + 1)} = \frac{	an 	heta}{\sec 	heta + 1}.

Recall 	an 	heta = \frac{\sin 	heta}{\cos 	heta}, \sec 	heta = \frac{1}{\cos 	heta}.

So,
\frac{	an 	heta}{\sec 	heta + 1} = \frac{\sin 	heta / \cos 	heta}{1/\cos 	heta + 1} = \frac{\sin 	heta / \cos 	heta}{(1 + \cos 	heta)/\cos 	heta} = \frac{\sin 	heta}{1 + \cos 	heta}.

Using the half-angle formula:
\frac{\sin 	heta}{1 + \cos 	heta} = 	an \frac{	heta}{2}.

Therefore,

$	an^{-1}\frac{\sqrt{1 + x^2} - 1}{x} = 	an^{-1}(	an \frac{	heta}{2}) = \frac{	heta}{2} = \frac{1}{2} 	an^{-1} x$.

4. Simplify $	an^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}ight), 0 < x < \pi$:

Note that:
\sqrt{\frac{1 - \cos x}{1 + \cos x}} = \sqrt{\frac{2 \sin^2 (x/2)}{2 \cos^2 (x/2)}} = \frac{\sin (x/2)}{\cos (x/2)} = 	an (x/2).

Therefore,

$	an^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}ight) = 	an^{-1}(	an (x/2)) = \frac{x}{2}$.

5. Simplify $	an^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}ight), \frac{-\pi}{4} < x < \frac{3\pi}{4}$:

Divide numerator and denominator by \cos x:

= 	an^{-1} \left( \frac{1 - 	an x}{1 + 	an x} ight).

Recall the tangent subtraction formula:
	an (\frac{\pi}{4} - x) = \frac{	an \frac{\pi}{4} - 	an x}{1 + 	an \frac{\pi}{4} 	an x} = \frac{1 - 	an x}{1 + 	an x}.

Therefore,

$	an^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}ight) = 	an^{-1}(	an (\frac{\pi}{4} - x)) = \frac{\pi}{4} - x$.

6. Simplify $	an^{-1}\frac{x}{\sqrt{a^2 - x^2}}, |x| < a$:

Let 	heta = \sin^{-1} \frac{x}{a}$, so that $x = a \sin 	heta$ and $	heta \in (-\pi/2, \pi/2)$.

Then,

$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sin^2 	heta} = a \cos 	heta$.

Therefore,

$\frac{x}{\sqrt{a^2 - x^2}} = \frac{a \sin 	heta}{a \cos 	heta} = 	an 	heta$.

Hence,

$	an^{-1} \frac{x}{\sqrt{a^2 - x^2}} = 	an^{-1} (	an 	heta) = 	heta = \sin^{-1} \frac{x}{a}$.

7. Simplify $	an^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}ight), a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$:

Note the formula for tangent of triple angle:

$	an 3\alpha = \frac{3 	an \alpha - 	an^3 \alpha}{1 - 3 	an^2 \alpha}$.

Let $t = \frac{x}{a}$, then the expression inside arctan is:

$\frac{3a^2 x - x^3}{a^3 - 3 a x^2} = \frac{3 a^3 t - a^3 t^3}{a^3 - 3 a^3 t^2} = \frac{3 t - t^3}{1 - 3 t^2} = 	an 3\alpha$ if $t = 	an \alpha$.

Since $\frac{-1}{\sqrt{3}} < t < \frac{1}{\sqrt{3}}$, $\alpha$ lies in $(-\pi/6, \pi/6)$.

Therefore,

$	an^{-1} \left( \frac{3a^2 x - x^3}{a^3 - 3 a x^2} ight) = 3 	an^{-1} \frac{x}{a}$. — Used trigonometric identities and substitutions:
- For Q3, used half-angle formula and algebraic manipulation.
- For Q4, used half-angle identity for cosine.
- For Q5, used tangent subtraction formula.
- For Q6, used substitution with sine inverse.
- For Q7, used triple angle formula for tangent.

Question 7

Find the values of each of the following:

8. $	an^{-1}\left[2\cos \left(2\sin^{-1}\frac{1}{2}ight)ight]$
9. $	an \frac{1}{2}\left[\sin^{-1}\frac{2x}{1 + x^2} + \cos^{-1}\frac{1 - y^2}{1 + y^2}ight], |x| < 1, y > 0 	ext{ and } xy < 1$

Accepted Answer

8. Evaluate $	an^{-1}\left[2\cos \left(2\sin^{-1}\frac{1}{2}ight)ight]$:

First, compute $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$.

Then,

$2 \sin^{-1} \frac{1}{2} = 2 	imes \frac{\pi}{6} = \frac{\pi}{3}$.

Calculate $\cos \frac{\pi}{3} = \frac{1}{2}$.

Therefore,

$2 \cos \left(2 \sin^{-1} \frac{1}{2} ight) = 2 	imes \frac{1}{2} = 1$.

Hence,

$	an^{-1} (1) = \frac{\pi}{4}$.

9. Evaluate $	an \frac{1}{2} \left[ \sin^{-1} \frac{2x}{1 + x^2} + \cos^{-1} \frac{1 - y^2}{1 + y^2} ight]$, where $|x| < 1, y > 0$ and $xy < 1$:

Note that:

$\sin^{-1} \frac{2x}{1 + x^2} = 2 	an^{-1} x$ (using the double angle formula for tangent).

Also,

$\cos^{-1} \frac{1 - y^2}{1 + y^2} = 2 	an^{-1} y$ (since $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$).

Therefore,

$\sin^{-1} \frac{2x}{1 + x^2} + \cos^{-1} \frac{1 - y^2}{1 + y^2} = 2 	an^{-1} x + 2 	an^{-1} y = 2 (	an^{-1} x + 	an^{-1} y)$.

Hence,

$	an \frac{1}{2} \left[ \sin^{-1} \frac{2x}{1 + x^2} + \cos^{-1} \frac{1 - y^2}{1 + y^2} ight] = 	an (	an^{-1} x + 	an^{-1} y)$.

Recall,

$	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.

So,

$	an (	an^{-1} x + 	an^{-1} y) = \frac{x + y}{1 - xy}$.

Therefore,

$\boxed{\frac{x + y}{1 - xy}}$. — For Q8, used inverse sine value and cosine evaluation.
For Q9, used double angle formulas for sine inverse and cosine inverse, then tangent addition formula.

Question 8

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

Accepted Answer

We know that $\sin^{-1}(\sin x) = x$ if $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Here, $\frac{2\pi}{3} = 120^\circ$ which is not in the principal range. Since $\sin \frac{2\pi}{3} = \sin \left(\pi - \frac{2\pi}{3}\right) = \sin \frac{\pi}{3}$, and $\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we have $\sin^{-1}\left(\sin \frac{2\pi}{3}\right) = \frac{\pi}{3}$. — The principal value of $\sin^{-1}$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Since $\frac{2\pi}{3}$ is outside this range, we use the identity $\sin x = \sin (\pi - x)$ to bring the angle into the principal range.

Question 9

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

Accepted Answer

The principal value of $\tan^{-1}$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Since $\frac{3\pi}{4} = 135^\circ$ is outside this range, we use the periodicity of tangent: $\tan x = \tan (x - \pi)$. Thus, $\tan \frac{3\pi}{4} = \tan \left(\frac{3\pi}{4} - \pi\right) = \tan \left(-\frac{\pi}{4}\right)$, and $-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Therefore, $\tan^{-1}\left(\tan \frac{3\pi}{4}\right) = -\frac{\pi}{4}$. — Since the principal branch of $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, angles outside this range are adjusted by subtracting or adding $\pi$ to bring them into the principal range.

Question 10

$\tan \left(\sin^{-1}\frac{3}{5} +\cot^{-1}\frac{3}{2}\right)$

Accepted Answer

Let $\alpha = \sin^{-1}\frac{3}{5}$ and $\beta = \cot^{-1}\frac{3}{2}$.

From $\alpha = \sin^{-1}\frac{3}{5}$, $\sin \alpha = \frac{3}{5}$.
Using Pythagoras, $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$.

From $\beta = \cot^{-1}\frac{3}{2}$, $\cot \beta = \frac{3}{2}$ implies $\tan \beta = \frac{2}{3}$.

We want $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.

Calculate $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$.

Therefore,
$$
\tan(\alpha + \beta) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}} = \frac{\frac{9}{12} + \frac{8}{12}}{1 - \frac{6}{12}} = \frac{\frac{17}{12}}{\frac{6}{12}} = \frac{17}{12} \times \frac{12}{6} = \frac{17}{6}.
$$

Hence, $\tan \left(\sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2}\right) = \frac{17}{6}$. — Use the definitions of inverse trigonometric functions to find sine, cosine, and tangent values, then apply the tangent addition formula.

Question 11

$\cos^{-1}\left(\cos \frac{7\pi}{6}\right)$  is equal to

(A)  $\frac{7\pi}{6}$

(B)  $\frac{5\pi}{6}$

(C)  $\frac{\pi}{3}$

(D)  $\frac{\pi}{6}$

Accepted Answer

The principal value of $\cos^{-1}$ lies in $[0, \pi]$. Since $\frac{7\pi}{6} > \pi$, we use the identity $\cos x = \cos (2\pi - x)$.

$\cos \frac{7\pi}{6} = \cos \left(2\pi - \frac{7\pi}{6}\right) = \cos \frac{5\pi}{6}$.

Since $\frac{5\pi}{6} \in [0, \pi]$,

$\cos^{-1}\left(\cos \frac{7\pi}{6}\right) = \frac{5\pi}{6}$.

Hence, correct option is (B). — The principal value of $\cos^{-1}$ is between 0 and $\pi$. Angles outside this range are adjusted using the cosine symmetry property.

Question 12

$\sin \left(\frac{\pi}{3} -\sin^{-1}(-\frac{1}{2})\right)$  is equal to

(A)  $\frac{1}{2}$

(B)  $\frac{1}{3}$

(C)  $\frac{1}{4}$

(D)  $1$

Accepted Answer

First, $\sin^{-1}(-\frac{1}{2}) = -\sin^{-1}\frac{1}{2} = -\frac{\pi}{6}$.

Therefore,
$$
\sin \left(\frac{\pi}{3} - \sin^{-1}(-\frac{1}{2})\right) = \sin \left(\frac{\pi}{3} - (-\frac{\pi}{6})\right) = \sin \left(\frac{\pi}{3} + \frac{\pi}{6}\right) = \sin \frac{\pi}{2} = 1.
$$

Hence, correct option is (D). — Use the property $\sin^{-1}(-x) = -\sin^{-1}(x)$ and then apply the sine addition formula.

Question 13

$\tan^{-1}\sqrt{3} -\cot^{-1}(-\sqrt{3})$  is equal to

(A)  $\pi$

(B)  $-\frac{\pi}{2}$

(C)  $0$

(D)  $2\sqrt{3}$

Accepted Answer

We have:
$$
\tan^{-1} \sqrt{3} = \frac{\pi}{3},
$$

and

$$
\cot^{-1}(-\sqrt{3}) = \pi - \tan^{-1}(-\sqrt{3}) = \pi + \tan^{-1} \sqrt{3} = \pi + \frac{\pi}{3} = \frac{4\pi}{3}.
$$

Since $\cot^{-1} x$ principal value lies in $(0, \pi)$, we adjust:
Actually, $\cot^{-1}(-\sqrt{3}) = \pi - \tan^{-1}(-\sqrt{3}) = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$ which is outside principal range, so principal value is $\cot^{-1}(-\sqrt{3}) = \frac{2\pi}{3}$ (since $\cot^{-1} x + \tan^{-1} x = \frac{\pi}{2}$ for positive x).

Alternatively, note:
$$
\cot^{-1}(-\sqrt{3}) = \pi - \tan^{-1}(-\sqrt{3}) = \pi + \frac{\pi}{3} = \frac{4\pi}{3},
$$
which is outside principal range $(0, \pi)$, so subtract $\pi$ to get $\frac{\pi}{3}$.

Hence,
$$
\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}.
$$

But none of the options match $-\frac{\pi}{3}$. Let's re-express carefully:

Since $\cot^{-1} y = \tan^{-1} \frac{1}{y}$ for $y \neq 0$,

$$
\cot^{-1}(-\sqrt{3}) = \tan^{-1} \left(-\frac{1}{\sqrt{3}}\right) = -\tan^{-1} \frac{1}{\sqrt{3}} = -\frac{\pi}{6}.
$$

Therefore,
$$
\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}.
$$

Since $\frac{\pi}{2}$ is not among options, check options again.

Options are:
(A) $\pi$
(B) $-\frac{\pi}{2}$
(C) $0$
(D) $2\sqrt{3}$

None is $\frac{\pi}{2}$. Possibly a typo in options or question.

Assuming the intended answer is $\frac{\pi}{2}$, closest is none.

Hence, the value is $\frac{\pi}{2}$.

Summary:
$$
\tan^{-1} \sqrt{3} = \frac{\pi}{3}, \quad \cot^{-1}(-\sqrt{3}) = -\frac{\pi}{6},
$$
so difference is $\frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}$. — Use the relation $\cot^{-1} x = \tan^{-1} \frac{1}{x}$ and carefully consider the signs and principal values of inverse functions.

Question 14

Find the value of the following:

1.  $\cos^{-1}\left(\cos \frac{13\pi}{6}\right)$
2.  $\tan^{-1}\left(\tan \frac{7\pi}{6}\right)$

Accepted Answer

1. To find $\cos^{-1}\left(\cos \frac{13\pi}{6}\right)$, first reduce $\frac{13\pi}{6}$ to an angle in the principal range of $\cos^{-1}$, which is $[0, \pi]$.

$\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$.

Since cosine is periodic with period $2\pi$,
$$ \cos \frac{13\pi}{6} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}. $$

Now, $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$ because $\frac{\pi}{6} \in [0, \pi]$.

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

2. To find $\tan^{-1}\left(\tan \frac{7\pi}{6}\right)$, note that the principal value of $\tan^{-1}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

$\frac{7\pi}{6} = \pi + \frac{\pi}{6} = 180^\circ + 30^\circ = 210^\circ$.

Since tangent has period $\pi$,
$$ \tan \frac{7\pi}{6} = \tan \left(\frac{7\pi}{6} - \pi\right) = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}. $$

But $\frac{\pi}{6} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for the original angle $\frac{7\pi}{6}$, so the principal value is $\tan^{-1}\left(\tan \frac{7\pi}{6}\right) = \tan^{-1}\left(\tan \frac{\pi}{6}\right) = \frac{\pi}{6}.$

However, since $\frac{7\pi}{6}$ is in the third quadrant where tangent is positive, the principal value is $-\frac{5\pi}{6}$ or $\frac{\pi}{6} - \pi = -\frac{5\pi}{6}$ which is not in the principal range.

Actually, the principal value of $\tan^{-1}(\tan \theta)$ is the unique angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with the same tangent.

Since $\tan \frac{7\pi}{6} = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$, the principal value is $\frac{\pi}{6}$.

Therefore, $\tan^{-1}\left(\tan \frac{7\pi}{6}\right) = \frac{\pi}{6}.$ — Step 1: Use periodicity and principal value ranges of inverse trig functions.
Step 2: Reduce angles to principal branches.
Step 3: Evaluate inverse trig functions accordingly.

Question 15

Prove that

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

Accepted Answer

Let $\theta = \sin^{-1} \frac{3}{5}$. Then $\sin \theta = \frac{3}{5}$.

Using Pythagoras theorem,
$$ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}. $$

Now,
$$ 2\sin^{-1} \frac{3}{5} = 2\theta. $$

Using double angle formula for tangent,
$$ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}. $$

Calculate $\tan \theta$:
$$ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}. $$

Therefore,
$$ \tan 2\theta = \frac{2 \times \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} = \frac{\frac{3}{2}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}. $$

Hence,
$$ 2\sin^{-1} \frac{3}{5} = 2\theta = \tan^{-1} \frac{24}{7}. $$ — Use substitution $\theta = \sin^{-1} \frac{3}{5}$, then apply double angle formula for tangent and simplify.

Mathematics

Mathematics — Study Notes

2.1 Introduction

2.2 Basic Concepts

2.2 Basic Concepts (continued) - Inverse of Cosecant, Secant, Tangent and Cotangent

Practice Questions — Mathematics

All 6 Chapters in Mathematics Part-I