Q.Find the principal value of the following: $\cos^{-1} \left(\cos \frac{13\pi}{6}\right)$

Compute $X$.

• Inner: $\sin^{-1}\tfrac12=\tfrac{\pi}{6}$, so $2\sin^{-1}\tfrac12=\tfrac{\pi}{3}$ and $\cos\tfrac{\pi}{3}=\tfrac12$. Then $2\cos(\cdots)=1$ and $\tan^{-1}(1)=\tfrac{\pi}{4}$.
• $\cos^{-1}\!\big[\cos\tfrac{7\pi}{6}\big]$: since $\tfrac{7\pi}{6}\notin[0,\pi]$ and $\cos\tfrac{7\pi}{6}=-\tfrac{\sqrt3}{2}$, we get $\cos^{-1}\!\big(-\tfrac{\sqrt3}{2}\big)=\tfrac{5\pi}{6}$.

Compute $Y$.

• $\sin^{-1}\!\big[\sin\tfrac{11\pi}{6}\big]$: $\sin\tfrac{11\pi}{6}=-\tfrac12$ and the principal value in $[-\tfrac\pi2,\tfrac\pi2]$ is $-\tfrac{\pi}{6}$.
• $\tan^{-1}\!\big[\tan\tfrac{4\pi}{3}\big]$: $\tan\tfrac{4\pi}{3}=\tan\tfrac{\pi}{3}=\sqrt3$, and the principal value is $\tfrac{\pi}{3}$.

Combine.

The problem asks for the sum

Each term is of the form $f^{-1}(f(\theta))$, but the result is not simply $\theta$ — it depends on whether $\theta$ lies in the principal‑value range of the inverse function. The ranges are:

• $\cos^{-1} x \in [0,\pi]$
• $\sin^{-1} x \in [-\pi/2,\pi/2]$
• $\tan^{-1} x \in (-\pi/2,\pi/2)$

So we must “adjust” each angle to an equivalent angle inside the respective range.

1. First term: $\cos^{-1}\!\left(\cos\frac{7\pi}{6}\right)$ $\frac{7\pi}{6} = 210^\circ$ is in the third quadrant, where cosine is negative. The principal range for $\cos^{-1}$ is $[0,\pi]$, so we need an angle in $[0,\pi]$ with the same cosine. Cosine is symmetric: $\cos(\pi + \theta) = -\cos\theta$, and $\cos(\pi - \theta) = -\cos\theta$. Here $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$, so $\cos\frac{7\pi}{6} = -\cos\frac{\pi}{6}$. The angle in $[0,\pi]$ with cosine $-\frac{\sqrt{3}}{2}$ is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$. Hence

2. Second term: $\sin^{-1}\!\left(\sin\frac{22\pi}{3}\right)$ First reduce $\frac{22\pi}{3}$ modulo $2\pi$: $\frac{22\pi}{3} = 6\pi + \frac{4\pi}{3}$ (since $6\pi = 3\cdot 2\pi$). So $\frac{22\pi}{3}$ is coterminal with $\frac{4\pi}{3}$. $\frac{4\pi}{3} = 240^\circ$ is in the third quadrant, where sine is negative. The principal range for $\sin^{-1}$ is $[-\pi/2,\pi/2]$. $\sin\frac{4\pi}{3} = -\sin\frac{\pi}{3}$. The angle in $[-\pi/2,\pi/2]$ with sine $-\frac{\sqrt{3}}{2}$ is $-\frac{\pi}{3}$. Therefore

3. Third term: $\tan^{-1}\!\left(\tan\frac{4\pi}{5}\right)$ $\frac{4\pi}{5} = 144^\circ$ is in the second quadrant, where tangent is negative. The principal range for $\tan^{-1}$ is $(-\pi/2,\pi/2)$. Tangent has period $\pi$: $\tan(\pi - \theta) = -\tan\theta$. Here $\frac{4\pi}{5} = \pi - \frac{\pi}{5}$, so $\tan\frac{4\pi}{5} = -\tan\frac{\pi}{5}$. The angle in $(-\pi/2,\pi/2)$ with tangent $-\tan\frac{\pi}{5}$ is $-\frac{\pi}{5}$. Hence

4. Sum them up:

Common denominator $30$:

The function $\tan^{-1}(\tan x)$ is not simply $x$ for all $x$ — it's the inverse of the tangent function only on the restricted domain where tangent is one-to-one. The standard principal branch of $\tan^{-1}$ gives outputs in $(-\pi/2, \pi/2)$. So when we feed $\tan(7\pi/6)$ into $\tan^{-1}$, we get back the unique angle in that interval whose tangent equals $\tan(7\pi/6)$.

Let’s work through it step by step.

1. Find the actual tangent value.

   The angle $\frac{7\pi}{6}$ is in the third quadrant (since $\pi < \frac{7\pi}{6} < \frac{3\pi}{2}$). In the third quadrant, tangent is positive.

   The reference angle is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$.

   So $\tan\left(\frac{7\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

2. Now apply the inverse tangent.

   We need $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

   The principal value of $\tan^{-1}$ lies in $(-\pi/2, \pi/2)$. The angle in that interval whose tangent is $\frac{1}{\sqrt{3}}$ is $\frac{\pi}{6}$.

3. Conclusion.

   Therefore, $\tan^{-1}\left(\tan \frac{7 \pi}{6}\right) = \frac{\pi}{6}$.

The principal value $-\tfrac{2\pi}{5}$ is negative, so the outer inverse function must be able to return a negative angle. The ranges of $\cos^{-1}$ and $\sec^{-1}$ are $[0,\pi]$ (never negative), which rules out (C) and (D).

Check (A): $\sin^{-1}$ has range $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$. Write

Since $\tfrac{2\pi}{5}\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$,

Check (B): $\tan$ has period $\pi$, so $\tan\tfrac{7\pi}{5}=\tan\tfrac{2\pi}{5}$, and $\tan^{-1}\!\big[\tan\tfrac{7\pi}{5}\big]=+\tfrac{2\pi}{5}$ — wrong sign.

The logarithm function $f(x) = \log_b x$ with base $b > 1$ is the inverse of the exponential function $g(x) = b^x$. Understanding this inverse relationship is the key to answering every part of the question.

Since $b^x$ is defined for all real $x$ and outputs only positive numbers, its inverse — the logarithm — must take only positive inputs and can output any real number. That immediately tells us the domain and range. Also, because $b^x$ is strictly increasing for $b > 1$, its inverse is also strictly increasing. And since $b^0 = 1$, the inverse must satisfy $\log_b 1 = 0$, so the point $(1, 0)$ is always on the graph.

Let's examine each option carefully.

1. Option (A): Domain is $\mathbb{R}$

   The logarithm $\log_b x$ is defined only for $x > 0$. You cannot take the log of zero or a negative number in the real number system. So the domain is $(0, \infty)$, not $\mathbb{R}$. This is false.

2. Option (B): Range is $\mathbb{R}^+$

   As $x \to 0^+$, $\log_b x \to -\infty$; as $x \to \infty$, $\log_b x \to \infty$. The logarithm can produce every real number, positive, negative, and zero. So the range is $\mathbb{R}$, not $\mathbb{R}^+$. This is false.

3. Option (C): The point $(1, 0)$ is always on the graph

   Since $b^0 = 1$ for any $b > 0$, the inverse function satisfies $\log_b 1 = 0$. Therefore, the graph of $y = \log_b x$ always passes through $(1, 0)$, regardless of the base (as long as $b > 0$ and $b \neq 1$). This is true.

4. Option (D): The graph is decreasing

   For $b > 1$, the exponential $b^x$ is increasing, so its inverse $\log_b x$ is also increasing. As $x$ increases, $\log_b x$ increases. The graph rises from left to right, not falls. This is false.

Evaluate each term using principal values.

• $\operatorname{cosec}^{-1}\!\left(-\dfrac{2\sqrt3}{3}\right)=\operatorname{cosec}^{-1}\!\left(-\dfrac{2}{\sqrt3}\right)$: $\sin\theta=-\dfrac{\sqrt3}{2}$ on $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ gives $\theta=-\dfrac{\pi}{3}$.
• $\tan^{-1}\!\left(-\dfrac{\sqrt3}{3}\right)=\tan^{-1}\!\left(-\dfrac{1}{\sqrt3}\right)=-\dfrac{\pi}{6}$.
• $\sec^{-1}2$: $\cos\theta=\dfrac{1}{2}$ gives $\theta=\dfrac{\pi}{3}$.
• $\cos^{-1}\!\left(-\dfrac{1}{2}\right)=\dfrac{2\pi}{3}$.
• $\sin^{-1}\!\left(\dfrac{\sqrt2}{2}\right)=\sin^{-1}\!\left(\dfrac{1}{\sqrt2}\right)=\dfrac{\pi}{4}$.

Add them (the last term is subtracted):

The $-\dfrac{\pi}{3}$ and $+\dfrac{\pi}{3}$ cancel, leaving

Concept and intuition

$\tan^{-1}(\tan x)=x$ only when $x\in(-\pi/2,\pi/2)$; otherwise it returns the unique angle in that interval sharing the same tangent value.

Step-by-step solution

1. Simplify $\alpha$. $\tan\dfrac{5\pi}{4} = \tan\left(\pi+\dfrac{\pi}{4}\right) = \tan\dfrac{\pi}{4}=1$ (tangent has period $\pi$). The angle in $(-\pi/2,\pi/2)$ with tangent 1 is $\pi/4$.

2. Simplify $\beta$. $\tan\dfrac{2\pi}{3} = \tan\left(\pi-\dfrac{\pi}{3}\right) = -\tan\dfrac{\pi}{3} = -\sqrt3$, so $-\tan\dfrac{2\pi}{3}=\sqrt3$.

3. Check each option.
   • (A) $3\alpha=\tfrac{3\pi}{4}$, $4\beta=\tfrac{4\pi}{3}$ — not equal.
   • (B) $\alpha-\beta = \tfrac{\pi}{4}-\tfrac{\pi}{3} = -\tfrac{\pi}{12}$ — not $\tfrac{7\pi}{12}$.
   • (C) $4\alpha = 4\cdot\tfrac{\pi}{4}=\pi$; $3\beta=3\cdot\tfrac{\pi}{3}=\pi$ — equal.
   • (D) $\alpha+\beta = \tfrac{\pi}{4}+\tfrac{\pi}{3}=\tfrac{7\pi}{12}$ — positive, not $-\tfrac{7\pi}{12}$.

Simplify the arguments: $-\dfrac{3}{\sqrt3}=-\sqrt3$ and $-\dfrac{2}{\sqrt2}=-\sqrt2$.

Combine:

We need to evaluate

The trap is that $\frac{35\pi}{18}$ is not inside the principal ranges of $\cos^{-1}$ or $\sin^{-1}$.

• For $\cos^{-1}$, the principal range is $[0, \pi]$.
• For $\sin^{-1}$, the principal range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

So we must first “unwrap” each function to find the equivalent angle inside those ranges.

1. Simplify $\cos^{-1}(\cos \frac{35\pi}{18})$ First, reduce $\frac{35\pi}{18}$ modulo $2\pi$:

because $2\pi = \frac{36\pi}{18}$.

Since $\cos(2\pi - \theta) = \cos\theta$, we have

Now $\frac{\pi}{18}$ is in $[0, \pi]$, so

2. Simplify $\sin^{-1}(\sin \frac{35\pi}{18})$ Again write $\frac{35\pi}{18} = 2\pi - \frac{\pi}{18}$. Since $\sin(2\pi - \theta) = -\sin\theta$, we get

Now $-\frac{\pi}{18}$ is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so

3. Subtract the two results