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Chapter 2 · Inverse Trigonometric Functions

Exercise 2.1 · Q14 (MCQ): Value of tan⁡−13−sec⁡−1(−2)\tan^{-1}\sqrt3-\sec^{-1}(-2)

Exercise 2.1Principal values · Section 2.2
The question

tan⁡−13−sec⁡−1(−2)\tan^{-1}\sqrt3 - \sec^{-1}(-2) is equal to

(A) π\pi
(B) −π3-\dfrac{\pi}{3}
(C) π3\dfrac{\pi}{3}
(D) 2π3\dfrac{2\pi}{3}

Exam answer

tan⁡−13=π3\tan^{-1}\sqrt3 = \frac{\pi}{3}⁠, as π3∈(−π2,π2)\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠.

Let sec⁡−1(−2)=y\sec^{-1}(-2) = y⁠. Then sec⁡y=−2\sec y = -2⁠, so cos⁡y=−12\cos y = -\frac12⁠, where y∈[0,π]−{π2}y \in [0, \pi] - \left\{\frac{\pi}{2}\right\}⁠. So y=π−π3=2π3y = \pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠.

tan⁡−13−sec⁡−1(−2)=π3−2π3=−π3\begin{aligned} &\tan^{-1}\sqrt3 - \sec^{-1}(-2) \\ &= \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \end{aligned}

Answer: (B) −π3-\dfrac{\pi}{3}

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Find the two principal values separately. tan⁡−13=π3\tan^{-1}\sqrt3 = \frac{\pi}{3}⁠. For sec⁡−1(−2)\sec^{-1}(-2)⁠, switch to cosine: sec⁡y=−2\sec y = -2 means cos⁡y=−12\cos y = -\frac12⁠, and in the range [0,π]−{π2}[0,\pi] - \left\{\frac{\pi}{2}\right\} that gives y=π−π3=2π3y = \pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠, not −π3-\frac{\pi}{3}⁠. The difference is π3−2π3=−π3\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}⁠.

Given, to find, and the plan
Given
tan⁡−13\tan^{-1}\sqrt3 (input 3∈R\sqrt3\in\R⁠) and sec⁡−1(−2)\sec^{-1}(-2) (input −2∈R−(−1,1)-2\in\R - (-1,1)⁠, since −2≤−1-2 \le -1⁠).
To find
The value of tan⁡−13−sec⁡−1(−2)\tan^{-1}\sqrt3 - \sec^{-1}(-2)⁠, and which option it matches.
Plan
(1) Find tan⁡−13\tan^{-1}\sqrt3 in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠. (2) Find sec⁡−1(−2)\sec^{-1}(-2) in [0,π]−{π2}[0,\pi] - \left\{\frac{\pi}{2}\right\}⁠. (3) Subtract in the given order. (4) Examine every option and the slip that produces it.

Basics you need

1Two principal ranges

tan⁡−1:R→(−π2,π2)\tan^{-1} : \R \to \left(-\frac{\pi}{2},\frac{\pi}{2}\right) and sec⁡−1:R−(−1,1)→[0,π]−{π2}\sec^{-1} : \R - (-1,1) \to [0,\pi] - \left\{\frac{\pi}{2}\right\}⁠. Lesson: principal values

2sec is 1/cos⁡1/\cos

sec⁡y=1cos⁡y\sec y = \frac{1}{\cos y}⁠, so sec⁡y=−2\sec y = -2 exactly when cos⁡y=−12\cos y = -\frac12⁠. Basics: tan, cot, sec and cosec

3Special angle π3\frac{\pi}{3}

sin⁡π3=32\sin\frac{\pi}{3} = \frac{\sqrt3}{2}⁠, cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠, so tan⁡π3=3\tan\frac{\pi}{3} = \sqrt3 and sec⁡π3=2\sec\frac{\pi}{3} = 2⁠. Basics: special angles and reference angles

4Allied angle

cos⁡(π−α)=−cos⁡α\cos(\pi - \alpha) = -\cos\alpha⁠, so also sec⁡(π−α)=−sec⁡α\sec(\pi - \alpha) = -\sec\alpha⁠. Basics: sin and cos on the unit circle

Solution

  1. First term: tan⁡−13\tan^{-1}\sqrt3

    Let a=tan⁡−13a = \tan^{-1}\sqrt3⁠. Then tan⁡a=3\tan a = \sqrt3 and −π2<a<π2-\frac{\pi}{2} < a < \frac{\pi}{2}⁠.

    Reference angle: tan⁡π3=3/21/2=3\tan\frac{\pi}{3} = \frac{\sqrt3/2}{1/2} = \sqrt3⁠, so α=π3\alpha = \frac{\pi}{3}⁠. Sign: 3>0\sqrt3 > 0⁠, and on (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) tan is positive exactly on (0,π2)\left(0,\frac{\pi}{2}\right)⁠, so

    a=tan⁡−13=π3.a = \tan^{-1}\sqrt3 = \frac{\pi}{3}.

    Exclusion: all solutions of tan⁡a=3\tan a = \sqrt3 are π3+nπ\frac{\pi}{3} + n\pi⁠; the neighbours −2π3-\frac{2\pi}{3} and 4π3\frac{4\pi}{3} are outside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠.

    Points with equal tan lie on one line through OO⁠. That line meets the circle at π3\frac{\pi}{3}⁠, on the right half (in range), and at the opposite point 4π3\frac{4\pi}{3}⁠, on the left half (left panel of the figure).
    tan⁡−1:R→(−π2,π2)\tan^{-1} : \R \to \left(-\frac{\pi}{2},\frac{\pi}{2}\right) (Section 2.2). Basics: tan, cot, sec and cosec

  2. Second term: sec⁡−1(−2)\sec^{-1}(-2)

    Let b=sec⁡−1(−2)b = \sec^{-1}(-2)⁠. Then sec⁡b=−2\sec b = -2⁠, with 0≤b≤π0 \le b \le \pi and b≠π2b \ne \frac{\pi}{2}⁠. Taking reciprocals (neither side is zero),

    cos⁡b=−12.\cos b = -\frac12.

    Reference angle: cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠, so α=π3\alpha = \frac{\pi}{3}⁠. Sign: on [0,π][0,\pi]⁠, cos is negative exactly on (π2,π]\left(\frac{\pi}{2},\pi\right]⁠, so b=π−αb = \pi - \alpha⁠:

    b=sec⁡−1(−2)=π−π3=2π3.b = \sec^{-1}(-2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}.

    Check: 2π3∈[0,π]\frac{2\pi}{3}\in[0,\pi]⁠, 2π3≠π2\frac{2\pi}{3}\ne\frac{\pi}{2} ✓; cos⁡2π3=−12\cos\frac{2\pi}{3} = -\frac12⁠, so sec⁡2π3=−2\sec\frac{2\pi}{3} = -2 ✓.

    Exclusion: all solutions of cos⁡b=−12\cos b = -\frac12 are ±2π3+2nπ\pm\frac{2\pi}{3} + 2n\pi⁠; the nearest others, −2π3-\frac{2\pi}{3} and 4π3\frac{4\pi}{3}⁠, are outside [0,π][0,\pi]⁠.

    The range of sec⁡−1\sec^{-1} is the range of cos⁡−1\cos^{-1} with π2\frac{\pi}{2} removed (where sec is undefined), so the sign rule is the cosine one: a negative input gives π−α\pi - \alpha⁠, never a negative angle.
    sec⁡−1:R−(−1,1)→[0,π]−{π2}\sec^{-1} : \R - (-1,1) \to [0,\pi] - \left\{\frac{\pi}{2}\right\} (Section 2.2). Lesson: principal values

  3. Subtract in the given order

    a−b=π3−2π3=−π3.\begin{aligned} a - b &= \frac{\pi}{3} - \frac{2\pi}{3}\\ &= -\frac{\pi}{3}. \end{aligned}
    O π/3 4π/3 −π/2 π/2 tan⁻¹ √3 = π/3 range (−π/2, π/2) O 2π/3 4π/3 0 π π/2 sec⁻¹(−2) = 2π/3 range [0, π] without π/2
    Left: tan⁡−13=π3\tan^{-1}\sqrt3 = \frac{\pi}{3} on the open right half (green). Right: sec⁡−1(−2)=2π3\sec^{-1}(-2) = \frac{2\pi}{3}⁠, the point of the upper half (with π2\frac{\pi}{2} removed) on the line x=−12x = -\frac12⁠. Rose points have the same ratio but lie outside the range.
    The answer is negative, and that is fine: a difference of two principal values is an ordinary number, not itself a principal value, so it need not lie in any principal range.
    π3−2π3=(1−2)π3\frac{\pi}{3} - \frac{2\pi}{3} = \frac{(1 - 2)\pi}{3}⁠. Basics: angles and radians

  4. Examine every option

    OptionValueVerdict
    (A)π\piWrong. It is what you get with sec⁡−1(−2)=−2π3\sec^{-1}(-2) = -\frac{2\pi}{3}⁠: π3+2π3=π\frac{\pi}{3} + \frac{2\pi}{3} = \pi⁠. But −2π3-\frac{2\pi}{3} is not in [0,π][0,\pi]⁠.
    (B)−π3-\frac{\pi}{3}Correct. π3−2π3=−π3\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} (Step 3).
    (C)π3\frac{\pi}{3}Wrong. It is b−ab - a⁠, the subtraction in the wrong order, or the result of ignoring the second term.
    (D)2π3\frac{2\pi}{3}Wrong. It is what you get with sec⁡−1(−2)=−π3\sec^{-1}(-2) = -\frac{\pi}{3}⁠, copying the sin⁡−1\sin^{-1} sign rule: π3+π3\frac{\pi}{3} + \frac{\pi}{3}⁠. But −π3-\frac{\pi}{3} is not in [0,π][0,\pi]⁠, and sec⁡(−π3)=+2\sec\left(-\frac{\pi}{3}\right) = +2⁠.
    Each wrong option corresponds to a specific slip, and every one of those slips uses an angle outside the principal range of sec⁡−1\sec^{-1}⁠, or reverses the subtraction. Checking the range of each term catches them all.
    For sec⁡−1\sec^{-1}⁠, as for cos⁡−1\cos^{-1}⁠: negative input gives π−α\pi - \alpha⁠. Lesson: principal values
Result

The correct answer is (B): tan⁡−13−sec⁡−1(−2)\tan^{-1}\sqrt3 - \sec^{-1}(-2) equals π3−2π3=−π3\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}⁠. (Textbook answer: B.)

Check it with numbers

In radians: tan⁡−1(1.7320508)≈1.0471976\tan^{-1}(1.7320508) \approx 1.0471976⁠. For the second term, sec⁡−1(−2)=cos⁡−1(−0.5)\sec^{-1}(-2) = \cos^{-1}(-0.5)⁠, which is ≈2.0943951\approx 2.0943951⁠. The difference is ≈−1.0471976\approx -1.0471976⁠, and −π3≈−1.0471976-\frac{\pi}{3} \approx -1.0471976 ✓.

Common mistakes
  1. Treating sec⁡−1\sec^{-1} like sin⁡−1\sin^{-1} and writing sec⁡−1(−2)=−π3\sec^{-1}(-2) = -\frac{\pi}{3}⁠. This leads to (D).
  2. Using a negative angle such as −2π3-\frac{2\pi}{3}, which has the right secant but lies outside [0,π][0,\pi]⁠. This leads to (A).
  3. Rejecting a negative answer because "angles from inverse trig functions are in the range". The individual terms are; their difference need not be.
What to take away
  1. Turn sec⁡−1\sec^{-1} into cos⁡−1\cos^{-1} by taking the reciprocal: sec⁡−1(−2)=cos⁡−1(−12)\sec^{-1}(-2) = \cos^{-1}\left(-\frac12\right)⁠.
  2. sec⁡−1\sec^{-1} follows the cosine sign rule: negative input gives π−α\pi - \alpha⁠.
  3. Evaluate each term in its own range, then combine; the combined result may be negative.

Every fact used, and where it comes from

FactUsed inSource
Principal ranges of tan⁡−1\tan^{-1} and sec⁡−1\sec^{-1}Steps 1, 2, 4Section 2.2 and its table of principal value branches (Lesson: principal values)
sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}Step 2Section 2.2 (definition of the sec function)
tan⁡π3=3\tan\frac{\pi}{3} = \sqrt3⁠, cos⁡π3=12\cos\frac{\pi}{3} = \frac12Steps 1, 2Class XI trigonometry (special angles)
cos⁡(π−α)=−cos⁡α\cos(\pi - \alpha) = -\cos\alpha⁠; periods of tan and cosSteps 1, 2Class XI trigonometry (allied angles)
cos⁡−1(−12)=2π3\cos^{-1}\left(-\frac12\right) = \frac{2\pi}{3}Step 2Same value as Exercise 2.1 Q5
All numerical valuesCheckRecomputed in the site's verification script
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