Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 12 Maths NCERT Solutions for Chapter 2 Ex 2.1 — all 14 questions solved, reading off the principal value of every inverse trigonometric function directly from its defined range.
Questions 1–10 each ask for the principal value of a single expression — sin⁻¹, cos⁻¹, cosec⁻¹, tan⁻¹, sec⁻¹ and cot⁻¹ all appear, so the real skill being tested is recalling each function's own principal value branch correctly (they aren't all the same interval — sec⁻¹ and cosec⁻¹ in particular trip students up). Questions 11 and 12 combine two or three of these principal values in a single sum, reusing results from earlier in the exercise. The exercise closes with two MCQs: one on the formal definition of sin⁻¹'s range, and one on evaluating a difference of two inverse trig values, tan⁻¹√3 − sec⁻¹(−2).
Let \sin^{-1}\left(-\frac12\right)=y. Then \sin y=-\frac12.
The range of the principal value branch of \sin^{-1} is \left[-\frac{\pi}{2},\frac{\pi}{2}\right], and \sin\left(-\frac{\pi}{6}\right)=-\frac12.
Let \cos^{-1}\left(\frac{\sqrt3}{2}\right)=y. Then \cos y=\frac{\sqrt3}{2}.
The range of the principal value branch of \cos^{-1} is [0,\pi], and \cos\frac{\pi}{6}=\frac{\sqrt3}{2}.
Let \text{cosec}^{-1}(2)=y. Then \text{cosec}\,y=2, i.e. \sin y=\frac12.
The range of the principal value branch of \text{cosec}^{-1} is \left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}, and \sin\frac{\pi}{6}=\frac12.
Let \tan^{-1}(-\sqrt3)=y. Then \tan y=-\sqrt3.
The range of the principal value branch of \tan^{-1} is \left(-\frac{\pi}{2},\frac{\pi}{2}\right), and \tan\left(-\frac{\pi}{3}\right)=-\sqrt3.
Let \cos^{-1}\left(-\frac12\right)=y. Then \cos y=-\frac12.
Since \cos\frac{2\pi}{3}=-\frac12 and \frac{2\pi}{3} lies in [0,\pi], the principal value branch of \cos^{-1}.
Let \tan^{-1}(-1)=y. Then \tan y=-1.
Since \tan\left(-\frac{\pi}{4}\right)=-1 and -\frac{\pi}{4} lies in \left(-\frac{\pi}{2},\frac{\pi}{2}\right), the principal value branch of \tan^{-1}.
Let \sec^{-1}\left(\frac{2}{\sqrt3}\right)=y. Then \sec y=\frac{2}{\sqrt3}, i.e. \cos y=\frac{\sqrt3}{2}.
The range of the principal value branch of \sec^{-1} is [0,\pi]-\left\{\frac{\pi}{2}\right\}, and \cos\frac{\pi}{6}=\frac{\sqrt3}{2}.
Let \cot^{-1}(\sqrt3)=y. Then \cot y=\sqrt3. The range of the principal value branch of \cot^{-1} is (0,\pi), and \cot\frac{\pi}{6}=\sqrt3.
Let \cos^{-1}\left(-\frac{1}{\sqrt2}\right)=y. Then \cos y=-\frac{1}{\sqrt2}.
Since \cos\frac{3\pi}{4}=-\frac{1}{\sqrt2} and \frac{3\pi}{4} lies in [0,\pi], the principal value branch of \cos^{-1}.
Let \text{cosec}^{-1}(-\sqrt2)=y. Then \text{cosec}\,y=-\sqrt2, i.e. \sin y=-\frac{1}{\sqrt2}.
Since \sin\left(-\frac{\pi}{4}\right)=-\frac{1}{\sqrt2} and -\frac{\pi}{4} lies in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}, the principal value branch of \text{cosec}^{-1}.
\tan^{-1}(1)=\dfrac{\pi}{4}; \cos^{-1}\left(-\frac12\right)=\dfrac{2\pi}{3} (Q5); \sin^{-1}\left(-\frac12\right)=-\dfrac{\pi}{6} (Q1).
Sum: \dfrac{\pi}{4}+\dfrac{2\pi}{3}-\dfrac{\pi}{6}=\dfrac{3\pi}{12}+\dfrac{8\pi}{12}-\dfrac{2\pi}{12}=\dfrac{9\pi}{12}.
\cos^{-1}\left(\frac12\right)=\dfrac{\pi}{3}; \sin^{-1}\left(\frac12\right)=\dfrac{\pi}{6}, so 2\sin^{-1}\left(\frac12\right)=\dfrac{\pi}{3}.
By definition, the principal value branch (range) of \sin^{-1} is the closed interval \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right] — the endpoints are included.
\tan^{-1}\sqrt3=\dfrac{\pi}{3}.
For \sec^{-1}(-2), we need y\in[0,\pi]-\left\{\frac{\pi}{2}\right\} with \sec y=-2, i.e. \cos y=-\frac12, giving y=\frac{2\pi}{3}.
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