📝 Inverse trig identities formulas (14 MCQs)

📖 Explanation: Since $\sin 0 = 0$ and the principal value of $\sin^{-1}$ returns the angle in $[-\,\frac{\pi}{2},\frac{\pi}{2}]$ whose sine is the argument, the only angle satisfying this is $0$. Hence $\sin^{-1}(0)=0$.

📖 Explanation: The fundamental identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$ holds for any $x$ in $[-1,1]$. Substituting $x=\frac12$ does not change the sum, so the result is $\frac{\pi}{2}$.

📖 Explanation: Let $\theta=\sin^{-1}x$; then $\sin\theta=x$ and $\cos\theta=\sqrt{1-x^{2}}$ (non‑negative). By definition $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{x}{\sqrt{1-x^{2}}}$. Hence the correct expression is $\frac{x}{\sqrt{1-x^{2}}}$.

📖 Explanation: The inverse sine returns the principal angle in $[-\,\frac{\pi}{2},\frac{\pi}{2}]$. Only when the original angle $x$ lies in that interval does $\sin^{-1}(\sin x)$ equal $x$; otherwise the value is adjusted by adding or subtracting multiples of $\pi$.

📖 Explanation: Apply the identity directly with $x=3$: $\sec(\tan^{-1}3)=\sqrt{1+3^{2}}=\sqrt{10}$. This avoids constructing a triangle; the formula already gives the required value.

📖 Explanation: Let $\theta=\sec^{-1}x$; then $\sec\theta=x$ and $\cos\theta=\frac{1}{x}$. By the Pythagorean theorem, $\sin\theta=\sqrt{1-\cos^{2}\theta}=\sqrt{1-\frac{1}{x^{2}}}=\frac{\sqrt{x^{2}-1}}{x}$.

📖 Explanation: Because $\theta$ is defined to be in $[-\,\frac{\pi}{2},\frac{\pi}{2}]$, its cosine is non‑negative. Using the identity $\sin^{2}\theta+\cos^{2}\theta=1$ and $\sin\theta=x$, we obtain $\cos\theta=\sqrt{1-x^{2}}$.

📖 Explanation: Since $x$ lies in $[\frac{\pi}{2},\frac{3\pi}{2}]$, the principal value is $-\,(x-\pi)=\pi-x$. Substituting $x=\frac{3\pi}{4}$ gives $\pi-\frac{3\pi}{4}=\frac{\pi}{4}$. Thus $\sin^{-1}(\sin x)=\frac{\pi}{4}$.

📖 Explanation: Starting with a right triangle where opposite side is $x$ and adjacent side is $1$, the hypotenuse is $\sqrt{1+x^{2}}$. The angle whose tangent is $x$ is $\tan^{-1}x$. The same angle has sine $\frac{x}{\sqrt{1+x^{2}}}$, so $\sin^{-1}$ of that ratio equals the same angle. Hence the two expressions are equal.

📖 Explanation: By definition, the inverse cosine returns the unique angle $\theta$ in $[0,\pi]$ whose cosine equals the given number. This interval ensures the function is one‑to‑one and covers all possible cosine values from $-1$ to $1$.

📖 Explanation: Compute $\frac{x}{\sqrt{1-x^{2}}}$ with $x=\frac{1}{\sqrt{2}}$: the denominator becomes $\sqrt{1-\frac12}=\frac{1}{\sqrt{2}}$, so the fraction equals $1$. Hence $\sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right)=\tan^{-1}(1)=\frac{\pi}{4}$.

📖 Explanation: From the identity $\alpha+\beta=\frac{\pi}{2}$, we can solve for $\alpha-\beta = 2\alpha-\frac{\pi}{2}$. Since $\alpha$ varies with $x$, the difference is not a constant; it changes according to the specific $x$ chosen.

📖 Explanation: Apply the formula with $x=-2$: $|x|=2$ and $\sqrt{x^{2}-1}=\sqrt{4-1}=\sqrt{3}$. The expression becomes $\frac{\sqrt{3}}{2}$, which is positive because the absolute value in the denominator removes the sign of $x$.

📖 Explanation: The correct form must handle both positive and negative $x$. The identity $\sin(\sec^{-1}x)=\frac{\sqrt{x^{2}-1}}{|x|}$ does exactly that, because the absolute value in the denominator ensures the result is always non‑negative, matching the range of the sine function for the principal angle.