The principal value of $\text{cosec}^{-1}(-2)$ is: (a) $\dfrac{-\pi}{3}$ (b) $\dfrac{-\pi}{6}$ (c) $\dfrac{\pi}{6}$ (d) $\dfrac{\pi}{3}$

The principal value of cosec⁻¹(-2) is: (a) (-π)/(3) (b) (-π)/(6) (c) (π)/(6) (d) (π)/(3)

Question 138 of 175

\text{cosec}^{-1}(-2)

(a)

\dfrac{-\pi}{3}

(b)

\dfrac{-\pi}{6}

(c)

\dfrac{\pi}{6}

(d)

\dfrac{\pi}{3}

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2018MCQ· 1mImportance★★★★★

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Concept understanding — Inverse Trigonometric Functions

Why an inverse needs a restricted domain. A function has an inverse only where it is one-to-one and onto. None of the six trigonometric functions is one-to-one on its natural domain, because each is periodic — infinitely many angles give the same sine, the same cosine, and so on. (Compare $y=x^2$ on all of $\mathbb R$ : it isn't one-to-one either, since $x=2$ and $x=-2$ both give $y=4$ ; only once the domain is cut down to $x\ge0$ does an inverse, $\sqrt x$ , exist.) So before " $\sin^{-1}$ ", " $\cos^{-1}$ ", etc. can mean anything, each trigonometric function is first restricted to one conventional interval on which it is one-to-one and whose image is its whole natural range. That restricted interval always contains $0$ together with some positive angles.

The six function/inverse domain-range pairs.

Function (restricted)	Domain $\to$ Range	Inverse	Domain $\to$ Range
$\sin x$	$\left[-\tfrac\pi2,\tfrac\pi2\right]\to[-1,1]$	$\sin^{-1}x$	$[-1,1]\to\left[-\tfrac\pi2,\tfrac\pi2\right]$
$\cos x$	$[0,\pi]\to[-1,1]$	$\cos^{-1}x$	$[-1,1]\to[0,\pi]$
$\tan x$	$\left(-\tfrac\pi2,\tfrac\pi2\right)\to(-\infty,\infty)$	$\tan^{-1}x$	$(-\infty,\infty)\to\left(-\tfrac\pi2,\tfrac\pi2\right)$
$\cot x$	$(0,\pi)\to(-\infty,\infty)$	$\cot^{-1}x$	$(-\infty,\infty)\to(0,\pi)$
$\csc x$	$\left[-\tfrac\pi2,\tfrac\pi2\right]-\{0\}\to\mathbb R-(-1,1)$	$\csc^{-1}x$	$\mathbb R-(-1,1)\to\left[-\tfrac\pi2,\tfrac\pi2\right]-\{0\}$
$\sec x$	$[0,\pi]-\left\{\tfrac\pi2\right\}\to\mathbb R-(-1,1)$	$\sec^{-1}x$	$\mathbb R-(-1,1)\to[0,\pi]-\left\{\tfrac\pi2\right\}$

Each inverse function's stated domain is exactly the range of the original restricted function, and its range is exactly that restricted domain — an inverse always swaps domain and range with the function it undoes, and its graph is the mirror image of the original's graph in the line $y=x$ . …

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Solving $\text{cosec}\,\theta=-2$ within the principal range $[-\pi/2,\pi/2]\setminus\{0\}$ gives $\theta=-\pi/6$ .

We need $\theta$ such that $\text{cosec}\,\theta = -2$ , i.e. $\sin\theta = -\dfrac{1}{2}$ , with $\theta$ in the principal value range of $\text{cosec}^{-1}$ , namely $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\setminus\{0\}$ .

Since $\sin\theta$ is negative, $\theta$ must be in the lower half of this range, i.e. $\theta\in\left[-\dfrac{\pi}{2},0\right)$ .

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Q.The principal value of cosec−1(−2)\text{cosec}^{-1}(-2)cosec−1(−2) is:

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Q.The principal value of $\text{cosec}^{-1}(-2)$ is:

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