Class 12 ncert solution math exercise 2.1

Embarking on the realm of inverse trigonometry in Class 12 NCERT solution Math Exercise 2.1 serve as an invaluable resource for students. This section dives into the intricate world of inverse trigonometric functions, unraveling their properties, and elucidating the nuances of solving problems related to them. The solutions provide a meticulous guide, offering step-by-step explanations to aid students in mastering the complexities of inverse trigonometry. Whether unraveling identities, understanding the domains, or solving equations, this resource is designed to enhance conceptual clarity.

     EXERCISE 2.1

Find the principal values of the following:(Class 12 ncert solution math exercise 2.1)

Question 1: \sin^{-1}(\frac{-1}{2})

Solution : Principal value of

\sin^{-1}(\frac{-1}{2})= -\frac{\pi}{6}\in[-\pi/2,\pi/2]

Question 2: \cos^{-1}(\frac{\sqrt{3}}{2})

Solution : Principal value of

\cos^{-1}(\frac{\sqrt{3}}{2})= \pi/6 \in[0, \pi]

Question 3: \operatorname{cosec}^{-1}(2)

Solution: Principal value of

\operatorname{cosec}^{-1}(2)= \pi/6\in[-pi/2, \pi/2]-\{0\}

Question 4: \tan^{-1}(-\sqrt{3})

Solution: Principal value of

\tan^{-1}(-\sqrt{3}) = -\pi/3\in(-\pi/2,\pi/2)

Question 5: \cos^{-1}(-\frac{1}{2})

Solution :Principal value of

\cos^{-1}(-\frac{1}{2})= \pi-\frac{\pi}{3}

= 2\pi/3 \in [0,\pi]

Question 6: \tan^{-1}(-1)

Solution: Principal value of

\tan^{-1}(-1) = -\pi/4\in(-\pi/2, \pi/2)

Question 7: \sec^{-1}(\frac{2}{\sqrt{3}})

Solution : Principal value of

\sec^{-1}(\frac{2}{\sqrt{3}})=\pi/6\in[0, \pi]-\{\pi/2\}

Question 8: \cot^{-1}(\sqrt{3})

Solution: Principal value of

\cot^{-1}(\sqrt{3}) = \pi/6\in(0,\pi)

Question 9: \cos^{-1}(-\frac{1}{\sqrt{2}})

Solution: Principal value of

\cos^{-1}(-\frac{1}{\sqrt{2}}) = \pi - \pi/4

= 3\pi/4\in (0, \pi)

Question 10: \operatorname{cosec}^{-1}(-\sqrt{2})

Solution: Principal value of

\operatorname{cosec}^{-1}(-\sqrt{2})=-\pi/4 \in [-\pi/2,\pi/2]-\{0\}

Find the values of the following:

Question 11: \tan^{-1}(1)+\cos^{-1}(-1/2)+\sin^{-1}(-1/2)

Solution: Since
\tan^{-1}(1) = \pi/4\in(-\pi/2,\pi/2)

\cos^{-1}(-1/2)= \pi - \pi/3

= 2\pi/3 \in[0,\pi]

\sin^{-1}(-1/2)=-\pi/6\in[-\pi/2,\pi/2]

Hence

\tan^{-1}(1) = \pi/4+2\pi/3-\pi/6

= \frac{3\pi+8\pi-2\pi}{12}

= \frac{9\pi}{12}

=\frac{3\pi}{12}

Question 12: \cos^{-1}(1/2)+2\sin^{-1}(1/2)

Solution : Since,

\cos^{-1}(1/2)= \pi/3\in[0, \pi]

\sin^{-1}(1/2)= \pi/6\in[-\pi/2, \pi/2]

Hence,
\cos^{-1}(1/2)+2\sin^{-1}(1/2) = \pi/3+2\pi/6

= 2\pi/3

Question 13: Find the value of \sin ^{-1} x=y, then

(A) 0 \leq y \leq \pi

(B) -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

(C) 0 \leq y \leq \pi

(D) -\frac{\pi}{2}<y<\frac{\pi}{2}

Answer: (B).

Question 14: Find the value of \tan ^{-1} \sqrt{3}-\sec ^{-1}(-2) is equal to

(A) 0

(B) -\frac{\pi}{3}

(C) \frac{\pi}{3}

(D) \frac{2 \pi}{3}

Answer (B)

Since, \tan ^{-1}(\sqrt{3})=\pi/3\in (-\pi/2, \pi/2)
Hence,

\sec^{-1}(-2)=2\pi/3\in[0,\pi]-\{\pi/2\}

Hence,

\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)= \pi/3-2\pi/3

= -\pi/3.


class 12 inverse trigonometric functions multiple choice

 

Case study inverse trigonometry 2
Two men on either side of a temple 30 metres high observes

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