Question

If $\sin^{-1}(\sin 4) + \cos^{-1}(\cos 8) + \tan^{-1}(\tan 6) + \cot^{-1}(\cot 10) = a + b\pi$ then (a + b) equals

• A. 26
• B. 14
• C. 12
• D. 10

Answer 1

Answer: (B)

14

Answer 2

The problem requires evaluating the sum of inverse trigonometric functions with arguments outside their principal ranges. We will use the properties of inverse trigonometric functions to find the equivalent values within the principal ranges. The principal ranges are:

$\sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$

$\cos^{-1}x \in [0, \pi]$

$\tan^{-1}x \in (-\frac{\pi}{2}, \frac{\pi}{2})$

$\cot^{-1}x \in (0, \pi)$

We use the approximation $\pi \approx 3.1416$.

$\pi/2 \approx 1.5708$

$\pi \approx 3.1416$

$3\pi/2 \approx 4.7124$

$2\pi \approx 6.2832$

$5\pi/2 \approx 7.8540$

$3\pi \approx 9.4248$

$7\pi/2 \approx 10.9956$

1. $\sin^{-1}(\sin 4)$:

The argument is 4. The principal range for $\sin^{-1}$ is $[-\pi/2, \pi/2]$.

Since $\pi < 4 < 3\pi/2$ (i.e., $3.1416 < 4 < 4.7124$), we use the property $\sin^{-1}(\sin x) = \pi - x$ for $x \in [\pi/2, 3\pi/2]$.

So, $\sin^{-1}(\sin 4) = \pi - 4$.

Check if $\pi - 4$ is in $[-\pi/2, \pi/2]$: $\pi - 4 \approx 3.1416 - 4 = -0.8584$.

Since $-1.5708 \le -0.8584 \le 1.5708$, the value is correct.

2. $\cos^{-1}(\cos 8)$:

The argument is 8. The principal range for $\cos^{-1}$ is $[0, \pi]$.

Since $2\pi < 8 < 3\pi$ (i.e., $6.2832 < 8 < 9.4248$), we use the property $\cos^{-1}(\cos x) = x - 2\pi$ for $x \in [2\pi, 3\pi]$.

So, $\cos^{-1}(\cos 8) = 8 - 2\pi$.

Check if $8 - 2\pi$ is in $[0, \pi]$: $8 - 2\pi \approx 8 - 6.2832 = 1.7168$.

Since $0 \le 1.7168 \le 3.1416$, the value is correct.

3. $\tan^{-1}(\tan 6)$:

The argument is 6. The principal range for $\tan^{-1}$ is $(-\pi/2, \pi/2)$.

Since $3\pi/2 < 6 < 2\pi$ (i.e., $4.7124 < 6 < 6.2832$), we use the property $\tan^{-1}(\tan x) = x - 2\pi$ for $x \in (3\pi/2, 5\pi/2)$.

So, $\tan^{-1}(\tan 6) = 6 - 2\pi$.

Check if $6 - 2\pi$ is in $(-\pi/2, \pi/2)$: $6 - 2\pi \approx 6 - 6.2832 = -0.2832$.

Since $-1.5708 < -0.2832 < 1.5708$, the value is correct.

4. $\cot^{-1}(\cot 10)$:

The argument is 10. The principal range for $\cot^{-1}$ is $(0, \pi)$.

Since $3\pi < 10 < 7\pi/2$ (i.e., $9.4248 < 10 < 10.9956$), we use the property $\cot^{-1}(\cot x) = x - 3\pi$ for $x \in (3\pi, 4\pi)$.

So, $\cot^{-1}(\cot 10) = 10 - 3\pi$.

Check if $10 - 3\pi$ is in $(0, \pi)$: $10 - 3\pi \approx 10 - 9.4248 = 0.5752$.

Since $0 < 0.5752 < 3.1416$, the value is correct.

Now, sum the terms:

$\sin^{-1}(\sin 4) + \cos^{-1}(\cos 8) + \tan^{-1}(\tan 6) + \cot^{-1}(\cot 10) = (\pi - 4) + (8 - 2\pi) + (6 - 2\pi) + (10 - 3\pi)$

$= (\pi - 2\pi - 2\pi - 3\pi) + (-4 + 8 + 6 + 10)$

$= (1 - 2 - 2 - 3)\pi + (-4 + 8 + 6 + 10)$

$= -6\pi + (4 + 16)$

$= -6\pi + 20$

$= 20 - 6\pi$

We are given that the sum equals $a + b\pi$.

Comparing $20 - 6\pi$ with $a + b\pi$, we have $a = 20$ and $b = -6$.

We need to find the value of $a+b$.

$a+b = 20 + (-6) = 20 - 6 = 14$.

The final answer is $\boxed{14}$.

Explanation of the solution:

1. Evaluate each term $\sin^{-1}(\sin 4)$, $\cos^{-1}(\cos 8)$, $\tan^{-1}(\tan 6)$, $\cot^{-1}(\cot 10)$ by finding the equivalent angle within the principal range of the respective inverse trigonometric function.
2. For $\sin^{-1}(\sin 4)$, since $\pi < 4 < 3\pi/2$, $\sin^{-1}(\sin 4) = \pi - 4$.
3. For $\cos^{-1}(\cos 8)$, since $2\pi < 8 < 3\pi$, $\cos^{-1}(\cos 8) = 8 - 2\pi$.
4. For $\tan^{-1}(\tan 6)$, since $3\pi/2 < 6 < 2\pi$, $\tan^{-1}(\tan 6) = 6 - 2\pi$.
5. For $\cot^{-1}(\cot 10)$, since $3\pi < 10 < 4\pi$, $\cot^{-1}(\cot 10) = 10 - 3\pi$.
6. Sum the evaluated terms: $(\pi - 4) + (8 - 2\pi) + (6 - 2\pi) + (10 - 3\pi) = (1 - 2 - 2 - 3)\pi + (-4 + 8 + 6 + 10) = -6\pi + 20 = 20 - 6\pi$.
7. Compare the result with $a + b\pi$ to find $a=20$ and $b=-6$.
8. Calculate $a+b = 20 + (-6) = 14$.

The final answer is $\boxed{14}$.