📝 How to evaluate inverse trig functions (16 MCQs)

📖 Explanation: The inverse sine is defined to be a function, so its output must be unique. By convention the range is limited to angles whose terminal side lies in the first or fourth quadrants, i.e., from $-\pi/2$ up to $\pi/2$. This ensures a one‑to‑one correspondence for all admissible $y$.

📖 Explanation: Applying the identity, we replace $x$ by $-3$ and obtain $\cos^{-1}\!\big(1/(-3)\big)=\cos^{-1}\!\left(-\frac{1}{3}\right)$. The other choices either ignore the reciprocal or use an incorrect sign, so option D is the only correct representation.

📖 Explanation: In the fourth quadrant the sine of an angle is negative, while cosine is positive. Since $\theta$ is the principal value of $\sin^{-1}$, the sine of $\theta$ equals $y$. Therefore $y$ must be negative, making option C the only logical conclusion.

📖 Explanation: The range of $\cos^{-1}$ is $[0,\pi]$. Angles in the second quadrant have measures between $\pi/2$ and $\pi$, where the cosine function is negative. Since $\cos(\alpha)=y$, the value of $y$ must be negative, so option B is correct.

📖 Explanation: The arctangent function is increasing on its domain. For angles between 0 and $\pi/4$ (approximately 0.785 rad), the tangent values lie strictly between 0 and 1. Hence $y$ must satisfy $0<y<1$, making option A the correct inference.

📖 Explanation: If $\sin^{-1}(y)=\pi/6$, then $y=\sin(\pi/6)=1/2$. Using the identity $\cos(\sin^{-1}t)=\sqrt{1-t^{2}}$ for $t$ in the principal range, we get $\cos(\pi/6)=\sqrt{1-(1/2)^{2}}=\sqrt{3}/2$. Thus option D is correct.

📖 Explanation: In the first quadrant ($0<\theta<\pi/2$) the cosine is positive, so its reciprocal, the secant, is also positive and has magnitude at least 1. Therefore $x$ must be greater than or equal to 1, making option B the only admissible interval.

📖 Explanation: For any $y$ in $[-1,1]$, the identity $\sin^{-1}y+\cos^{-1}y=\pi/2$ holds because the angles are complementary in a right triangle. Substituting $y=0.5$ gives the same relationship, so option C accurately reflects the connection.

📖 Explanation: Using the identity $\sin^{-1}t+\cos^{-1}t=\pi/2$ for any admissible $t$, the sum evaluates directly to $\pi/2$. No approximation is needed, and none of the other options satisfy the identity, so option B is correct.

📖 Explanation: The angle whose tangent equals $1/\sqrt{3}$ is $\pi/6$ because $\tan(\pi/6)=1/\sqrt{3}$. This is a standard special‑angle value, making option C the correct equivalence.

📖 Explanation: The arcsine function grows faster than the arctangent on $(0,1)$. Numerically, $\sin^{-1}(0.8)\approx0.927$ rad while $\tan^{-1}(0.8)\approx0.688$ rad, so the arcsine value is larger. Hence option A is the correct comparison.

📖 Explanation: If $\sec^{-1}(x)=\pi/3$, then $\sec(\pi/3)=x$. Since $\cos(\pi/3)=1/2$, its reciprocal is $2$. Therefore $x=2$, which corresponds to option D.

📖 Explanation: From the right‑triangle definition, if $\sin\theta=y$ then the adjacent side is $\sqrt{1-y^{2}}$. Thus $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{y}{\sqrt{1-y^{2}}}$. This matches option A.

📖 Explanation: Since $\sin^{2}\phi+\cos^{2}\phi=1$, we have $\cos\phi=\sqrt{1-(3/5)^{2}}=\sqrt{16/25}=4/5$. Because $\phi$ lies in the range $[0,\pi]$, the cosine is positive, giving $y=4/5$. Hence option D is correct.

📖 Explanation: The principal value of $\sin^{-1}$ is restricted to $[-\,\pi/2,\pi/2]$. Since the computed value is positive and clearly less than $\pi/2$, the angle must lie in the first quadrant, where both sine and cosine are positive. Option A captures this reasoning.

📖 Explanation: For any real angle, $|\cos\theta|\le1$. Taking the reciprocal gives $|\sec\theta|=1/|\cos\theta|\ge1$. Consequently, no real angle yields a secant whose absolute value is less than 1, so \(\