Inverse trigonometry practice · Medium

Option B ($\left[-1, \frac{{1}}{{\sqrt{{2}}}}\right]$) is correct because solving the inequality $\sin^{-1}x \le \cos^{-1}x$ leads to this interval.

Option A ($\frac{1}{\sqrt{2}}$) is correct because by using the formula for $\tan(A+B)$ and then finding the cosine, we arrive at the correct value.

Option D ($\frac{{25 \pi}}{{12}}$) is correct because it correctly evaluates each inverse trigonometric function and performs the arithmetic.

Option C ($\frac{\sqrt{17}-3}{4}$) is correct because solving the equation $\tan^{-1} x + \tan^{-1} (2x) = \frac{\pi}{4}$ leads to a quadratic equation, and the positive solution is $\frac{\sqrt{17}-3}{4}$.

Option D (11) is correct because by converting $\sec^{-1}2$ to $\tan^{-1}(\sqrt{3})$ and $\csc^{-1}3$ to $\cot^{-1}(2\sqrt{2})$, and then squaring, the result is 11.

Option B ($\frac{2 b}{a}$) is correct because simplifying the given expression using trigonometric identities leads to $\frac{2b}{a}$.

Option B ($\frac{\pi}{4}$) is correct because the expression simplifies to $\tan^{-1}(1)$, which equals $\frac{\pi}{4}$, using the inverse tangent subtraction formula.

Option D ($\frac{\pi}{12}$) is correct because by solving the given conditions, we find that $\alpha = \frac{\pi}{12}$.

Option B is correct because the assertion is correct, the reason is correct, but the reason is not a correct explanation for the assertion. The assertion is true because tangent is bijective on each interval, and the reason correctly states the principal value branch.

Option A ($\left[\frac{{1}}{{\sqrt{{2}}}}, 1\right]$) is correct because solving the given equation using trigonometric substitutions and considering the range of the inverse trigonometric functions leads to this interval.

Option B ($\frac{1+2 \sqrt{6}}{6}$) is correct because using the sine addition formula and the given values, the expression simplifies to this result.

Option C ($\frac{\pi }{4}$) is correct because by using the Pythagorean theorem and the tangent addition formula, we can simplify the expression to ${\tan ^{ - 1}}(1)$, which equals $\frac{\pi }{4}$.

Option A (2.36) is correct because simplifying the expression using trigonometric substitutions and identities leads to $\frac{{3\pi}}{{4}}$ which is approximately 2.36.

Option C ($\pm \sqrt{{\sin 2 \beta}}$) is correct because it's derived by applying componendo and dividendo and trigonometric identities to the given equation.

Option C ($\pi$) is correct because it utilizes the tangent addition formula and the given condition $x+y+z=xyz$ to simplify the expression.

Option A ($n$) is correct because each $\cos^{-1} \alpha_i$ must be 0, implying each $\alpha_i$ is 1, and their sum is $n$.

Option D ($(-\infty,-1] \cup[1, \infty)$) is correct because it represents the domain where both $\sec^{-1}x$ and $\tan^{-1}x$ are real.

Option B ($\frac{{\pi}}{{2}}$) is correct because simplifying the expression using trigonometric identities leads to $\frac{{\pi}}{{2}}$.

Option A (2 $abc$) is correct because the given condition ${\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi$ leads to the identity $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$, which simplifies to $a\sqrt {\left( {1 - {a^2}} \right)} + b\sqrt {\left( {1 - {b^2}} \right)} + c\sqrt {1 - {c^2})} = 2abc$.

Option B ($\frac{{-n^2}}{{(n^2+1)}}$) is correct because the given expression telescopes into a simpler form, allowing for easy differentiation and evaluation at x=0.

Option C ($3 \pi / 4$) is correct because the expression simplifies to $\pi + \tan^{-1}(-1)$, which equals $3\pi/4$ given the condition x > y > 0.

Option D (3) is correct because using the identity $\sin^{-1}x + \csc^{-1}y = \frac{{\pi}}{{2}}$ and solving for x yields the value 3.

Option C (0) is correct because the terms cancel out based on the given range of alpha and properties of inverse trigonometric functions.

Option B (1) is correct because solving the inequality for arctangent and then finding the largest integer value of x that satisfies the condition.

Option C (-7/17) is correct because it is the result of applying the tangent addition and double angle formulas to the given expression.

Option B (2n) is correct because the sum of inverse sines equaling nπ implies each term must be π/2, leading to each x_i being 1.

Option A (${\tan ^{ - 1}}\left( {\frac{{{{{n^2} + n}}}}{{{{{n^2} + n + 2}}}}} \right)$) is correct because the sum telescopes, leaving a simplified expression involving $n$.

Option C (3/4) is correct because by simplifying the given expression using trigonometric identities, we arrive at the value 3/4.

The correct answer is D ($\frac{\pi}{6}$) because it involves simplifying the expression using trigonometric identities and inverse trigonometric functions.

Option C (π/4) is correct because simplifying the expression inside out leads to sec⁻¹(cosec(cot(cot⁻¹(3π/4)))) = sec⁻¹(cosec(3π/4)) = sec⁻¹(√2) = π/4.