Inverse trigonometry practice · Mixed difficulty

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 C ($\frac{\pi}{6}$) is correct because the expression simplifies to $\frac{\pi}{6}$ after applying trigonometric identities and inverse trigonometric properties.

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 A ($\frac{{26}}{{25}}$) is correct because the telescoping sum simplifies to $\cot \left( {{\tan }^{ - 1}}\left( {50} \right) - {{\tan }^{ - 1}}\left( 1 \right) \right)$, which evaluates to $\frac{{26}}{{25}}$.

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 D ($\sqrt 3$) is correct because it is the solution obtained after simplifying the given trigonometric equation using inverse trigonometric identities.

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 D (none of these) is correct because the sum of the two inverse tangent functions simplifies to $\tan^{-1}(17/65)$, which is not equal to any of the other options.

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 ($x\sqrt {1 + {x^2}}$) is correct because simplifying the expression using trigonometric identities leads to this result.

Option C ($\mathrm{[sin\;}1,\mathrm{\;}1]$) is the correct solution set because it satisfies the given inequality $[{\sin ^{ - 1}}x] > [{\cos ^{ - 1}}x]$ considering the greatest integer function.

Option B ($\sqrt{5} / 3$) is correct because by using the trigonometric identity $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$, we can find the value of x.

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 B (Unique solution) is correct because solving the given equation using trigonometric identities results in a single valid solution for x.

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 D ($\frac{\pi }{2}$) is correct because if ${\sec ^{ - 1}}x = \mathrm{cose}{\mathrm{c}^{ - 1}}y$, then ${\cos ^{ - 1}}(\frac{1}{x}) = {\sin ^{ - 1}}(\frac{1}{y})$, and ${\cos ^{ - 1}}(\frac{1}{y}) + {\sin ^{ - 1}}(\frac{1}{y}) = \frac{\pi }{2}$.

Option C ($\pm \frac{2}{3}$) is correct because using the identity $\cos(2\theta) = 1 - 2\sin^2(\theta)$ and solving for $x = \sin(\theta)$ gives $\pm \frac{2}{3}$.

Option D ($\frac{\pi}{4}$) is correct because the expression simplifies to arctangent of 1, which equals $\frac{\pi}{4}$.

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

Option D ($- 1 \le x < \frac{1}{{\sqrt 2 }}$) is correct because, given the condition ${\cos ^{ - 1}}x > {\sin ^{ - 1}}x$, the solution lies in the interval $- 1 \le x < \frac{1}{{\sqrt 2 }}$.

Option C ($\frac{\pi }{4}$) is correct because by using the property that the sum of angles in a triangle is $\pi$ and the formula for $\tan^{-1}x + \tan^{-1}y$, we can solve for angle C.

Option B ($\frac{1}{{\sqrt {10} }}$) is correct because using the half-angle formula for sine, the expression simplifies to $\sqrt{\frac{1 - \cos \theta}{2}} = \frac{1}{\sqrt{10}}$.

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 A ($\alpha$) is correct because, after simplifying the given expression and solving for $\alpha$, the principal value of ${\sin ^{ - 1}}\sin \left( {\frac{{2\pi }}{3}} \right)$ is equal to $\alpha$.

Option B is correct because solving the equation ${\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \frac{\pi }{6}$ using trigonometric identities leads to $x = \frac{{\sqrt 3 }}{2}$.

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.

The correct answer is D (None of these) because the value of the expression is 0.28, which is not among the given options.