Question

Correct statement(s) is/are:

$\boxed{\phantom{x}}$ $cos^{-1}(\frac{12}{13}) + sin^{-1}(\frac{3}{5}) = sin^{-1}(\frac{56}{65})$

$\boxed{\phantom{x}}$ $tan^{-1}(\frac{cos x}{1 - sin x}) = \frac{\pi}{4} + \frac{x}{2}$ for $\frac{-3\pi}{2} < x < \frac{\pi}{2}$

$\boxed{\phantom{x}}$ $cot(\sum_{n=1}^{23} cot^{-1}(1 + \sum_{k=1}^{n} 2k)) = \frac{23}{25}$

$\boxed{\phantom{x}}$ $sin^{-1}(sin 8) + cos^{-1}(cos 15) = 7 - \pi$

• A. $cos^{-1}(\frac{12}{13}) + sin^{-1}(\frac{3}{5}) = sin^{-1}(\frac{56}{65})$
• B. $tan^{-1}(\frac{cos x}{1 - sin x}) = \frac{\pi}{4} + \frac{x}{2}$ for $\frac{-3\pi}{2} < x < \frac{\pi}{2}$
• C. $cot(\sum_{n=1}^{23} cot^{-1}(1 + \sum_{k=1}^{n} 2k)) = \frac{23}{25}$
• D. $sin^{-1}(sin 8) + cos^{-1}(cos 15) = 7 - \pi$

Answer 1

Answer: (A), (B), (D)

Statements 1, 2, and 4 are correct.

Answer 2

Let's evaluate each statement.

Statement 1: $cos^{-1}(\frac{12}{13}) + sin^{-1}(\frac{3}{5}) = sin^{-1}(\frac{56}{65})$

Let $A = cos^{-1}(\frac{12}{13})$ and $B = sin^{-1}(\frac{3}{5})$. Since $\frac{12}{13} > 0$ and $\frac{3}{5} > 0$, both $A$ and $B$ are in the interval $(0, \frac{\pi}{2})$. $cos A = \frac{12}{13} \implies sin A = \sqrt{1 - (\frac{12}{13})^2} = \frac{5}{13}$ (since $A \in (0, \frac{\pi}{2})$). $sin B = \frac{3}{5} \implies cos B = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$ (since $B \in (0, \frac{\pi}{2})$). Consider $sin(A+B) = sin A cos B + cos A sin B = (\frac{5}{13})(\frac{4}{5}) + (\frac{12}{13})(\frac{3}{5}) = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}$. Since $A, B \in (0, \frac{\pi}{2})$, $A+B \in (0, \pi)$. Also, $A = cos^{-1}(\frac{12}{13}) < cos^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}$ and $B = sin^{-1}(\frac{3}{5}) < sin^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}$. Thus, $A+B < \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$. Since $A+B \in (0, \frac{\pi}{2})$ and $sin(A+B) = \frac{56}{65}$, we have $A+B = sin^{-1}(\frac{56}{65})$. So, $cos^{-1}(\frac{12}{13}) + sin^{-1}(\frac{3}{5}) = sin^{-1}(\frac{56}{65})$ is correct.

Statement 2: $tan^{-1}(\frac{cos x}{1 - sin x}) = \frac{\pi}{4} + \frac{x}{2}$ for $\frac{-3\pi}{2} < x < \frac{\pi}{2}$.

Consider the expression inside the $tan^{-1}$: $\frac{cos x}{1 - sin x} = \frac{cos^2(x/2) - sin^2(x/2)}{cos^2(x/2) + sin^2(x/2) - 2 sin(x/2) cos(x/2)} = \frac{(cos(x/2) - sin(x/2))(cos(x/2) + sin(x/2))}{(cos(x/2) - sin(x/2))^2}$ Assuming $cos(x/2) - sin(x/2) \neq 0$, this simplifies to $\frac{cos(x/2) + sin(x/2)}{cos(x/2) - sin(x/2)}$. Dividing the numerator and denominator by $cos(x/2)$ (assuming $cos(x/2) \neq 0$), we get $\frac{1 + tan(x/2)}{1 - tan(x/2)} = tan(\frac{\pi}{4} + \frac{x}{2})$. So, $tan^{-1}(\frac{cos x}{1 - sin x}) = tan^{-1}(tan(\frac{\pi}{4} + \frac{x}{2}))$. For $tan^{-1}(tan \theta) = \theta$, we need $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Let $\theta = \frac{\pi}{4} + \frac{x}{2}$. The given interval for $x$ is $\frac{-3\pi}{2} < x < \frac{\pi}{2}$. Dividing by 2: $\frac{-3\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$. Adding $\frac{\pi}{4}$: $\frac{-3\pi}{4} + \frac{\pi}{4} < \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{4} + \frac{\pi}{4}$. $\frac{-\pi}{2} < \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{2}$. Since $\frac{\pi}{4} + \frac{x}{2}$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$, $tan^{-1}(tan(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$. The conditions $cos(x/2) - sin(x/2) \neq 0$ and $cos(x/2) \neq 0$ are satisfied in the interval $\frac{-3\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$. So, $tan^{-1}(\frac{cos x}{1 - sin x}) = \frac{\pi}{4} + \frac{x}{2}$ is correct.

Statement 3: $cot(\sum_{n=1}^{23} cot^{-1}(1 + \sum_{k=1}^{n} 2k)) = \frac{23}{25}$

The inner sum is $\sum_{k=1}^{n} 2k = 2 \sum_{k=1}^{n} k = 2 \frac{n(n+1)}{2} = n(n+1)$. The expression inside the $cot^{-1}$ is $1 + n(n+1)$. We use the identity $cot^{-1}(1 + n(n+1)) = tan^{-1}(\frac{1}{1 + n(n+1)})$. We can write $\frac{1}{1 + n(n+1)} = \frac{(n+1) - n}{1 + n(n+1)} = tan^{-1}(n+1) - tan^{-1}(n)$. The sum is $\sum_{n=1}^{23} (tan^{-1}(n+1) - tan^{-1}(n))$. This is a telescoping sum. The sum equals $(tan^{-1}(2) - tan^{-1}(1)) + (tan^{-1}(3) - tan^{-1}(2)) + \dots + (tan^{-1}(24) - tan^{-1}(23)) = tan^{-1}(24) - tan^{-1}(1)$. Since $tan^{-1}(1) = \frac{\pi}{4}$, the sum is $tan^{-1}(24) - \frac{\pi}{4}$. We need to evaluate $cot(tan^{-1}(24) - \frac{\pi}{4})$. Let $A = tan^{-1}(24)$. We need $cot(A - \frac{\pi}{4})$. $cot(A - \frac{\pi}{4}) = \frac{cot A cot(\frac{\pi}{4}) + 1}{cot(\frac{\pi}{4}) - cot A}$. Since $tan A = 24$, $cot A = \frac{1}{24}$. $cot(\frac{\pi}{4}) = 1$. $cot(A - \frac{\pi}{4}) = \frac{(\frac{1}{24})(1) + 1}{1 - \frac{1}{24}} = \frac{\frac{1+24}{24}}{\frac{24-1}{24}} = \frac{\frac{25}{24}}{\frac{23}{24}} = \frac{25}{23}$. The statement claims the value is $\frac{23}{25}$. This is incorrect.

Statement 4: $sin^{-1}(sin 8) + cos^{-1}(cos 15) = 7 - \pi$

For $sin^{-1}(sin x)$, we find the value $x' \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $sin x' = sin x$. $8$ radians is between $2\pi \approx 6.28$ and $3\pi \approx 9.42$. $sin 8 = sin(8 - 2\pi)$. $8 - 2\pi \approx 8 - 6.28 = 1.72$. $\frac{\pi}{2} \approx 1.57$. So $8 - 2\pi > \frac{\pi}{2}$. We use $sin x = sin(\pi - x)$. $sin(8 - 2\pi) = sin(\pi - (8 - 2\pi)) = sin(3\pi - 8)$. $3\pi - 8 \approx 9.42 - 8 = 1.42$. Since $1.42 \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $sin^{-1}(sin 8) = 3\pi - 8$.

For $cos^{-1}(cos x)$, we find the value $x' \in [0, \pi]$ such that $cos x' = cos x$. $15$ radians is between $4\pi \approx 12.56$ and $5\pi \approx 15.7$. $cos 15 = cos(15 - 4\pi)$. $15 - 4\pi \approx 15 - 12.56 = 2.44$. Since $2.44 \in [0, \pi]$, $cos^{-1}(cos 15) = 15 - 4\pi$.

The left side of the statement is $sin^{-1}(sin 8) + cos^{-1}(cos 15) = (3\pi - 8) + (15 - 4\pi) = 15 - 8 + 3\pi - 4\pi = 7 - \pi$. The right side of the statement is $7 - \pi$. So, $sin^{-1}(sin 8) + cos^{-1}(cos 15) = 7 - \pi$ is correct.

The correct statements are 1, 2, and 4.