Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.3 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.3

Question 1.

Find the principal values of the following :

(i) sin^{-1}\(\left(\frac{1}{2}\right)\)

Solution:

The principal value branch of sin^{-1}x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Let sin^{-1}\(\left(\frac{1}{2}\right)\) = ∝, where \(\frac{-\pi}{2}\) ≤ ∝ ≤ \(\frac{\pi}{2}\)

∴ sin∝ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\)

∴ ∝ = \(\frac{\pi}{6}\) …[∵ – \(\frac{\pi}{2}\) ≤ \(\frac{\pi}{6}\) ≤ \(\frac{\pi}{2}\)]

∴ the principal value of sin^{-1}\(\left(\frac{1}{2}\right)\) is \(\frac{\pi}{6}\).

(ii) cosec^{-1}(2)

Solution:

The principal value branch of cosec^{-1}x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.

Let cosec^{-1}(2) = ∝, where \(\frac{-\pi}{2}\) ≤ ∝ ≤ \(\frac{\pi}{2}\), ∝ ≠ 0

∴ cosec^{-1} ∝ = 2 = cosec\(\frac{\pi}{6}\)

∴ ∝ = \(\frac{\pi}{6}\) …[∵ –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{6}\) ≤ \(\frac{\pi}{2}\)]

∴ the principal value of cosec^{-1}(2) is \(\frac{\pi}{6}\).

(iii) tan^{-1}(-1)

Solution:

The principal value branch of tan^{-1}x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Let tan^{-1}(-1) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)

∴ tan∝ = -1 = -tan\(\frac{\pi}{4}\)

∴ tan∝ = tan\(\left(-\frac{\pi}{4}\right)\) …[∵ tan(-θ) = -tanθ]

∴ ∝ = –\(\frac{\pi}{4}\) …[∵ –\(\frac{\pi}{2}\) < \(\frac{-\pi}{4}\) < \(\frac{\pi}{2}\)]

∴ the principal value of tan^{-1}(-1) is –\(\frac{\pi}{4}\).

(iv) tan^{-1}(-\(\sqrt {3}\))

Solution:

The principal value branch of tan^{-1}x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

Let tan^{-1}(-\(\sqrt {3}\)) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)

∴ tan∝ = –\(\sqrt {3}\) = -tan\(\frac{\pi}{3}\)

∴ tan∝ = tan\(\left(-\frac{\pi}{3}\right)\) …[∵ tan(-θ) = -tanθ]

∴ ∝ = –\(\frac{\pi}{3}\) …[∵ –\(\frac{\pi}{2}\) < \(\frac{-\pi}{3}\) < \(\frac{\pi}{2}\)]

∴ the principal value of tan^{-1}(-\(\sqrt {3}\)) is –\(\frac{\pi}{3}\).

(v) sin^{-1} \(\left(\frac{1}{\sqrt{2}}\right)\)

Solution:

The principal value branch of sin^{-1}x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Let sin^{-1} \(\left(\frac{1}{\sqrt{2}}\right)\) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)

∴ sin∝ = \(\left(\frac{1}{\sqrt{2}}\right)\) = sin\(\frac{\pi}{4}\)

∴ ∝ = \(\frac{\pi}{4}\) …[∵ –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{4}\) ≤ \(\frac{\pi}{2}\)]

∴ the principal value of sin^{-1} \(\left(\frac{1}{\sqrt{2}}\right)\) is \(\frac{\pi}{4}\).

(vi) cos^{-1}\(\left(-\frac{1}{2}\right)\)

Solution:

The principal value branch of cos^{-1}x is (0, π).

Let cos^{-1}\(\left(-\frac{1}{2}\right)\) = ∝, where 0 ≤ ∝ ≤ π

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

∴ cos∝ = cos\(\left(\pi-\frac{\pi}{3}\right)\) …[∵ cos(π – θ) = -cosθ)

∴ cos∝ = cos\(\frac{2 \pi}{3}\)

∴ ∝ = \(\frac{2 \pi}{3}\) …[∵ 0 ≤ \(\frac{2 \pi}{3}\) ≤ π]

∴ the principal value of cos^{-1}\(\left(-\frac{1}{2}\right)\) is \(\frac{2 \pi}{3}\).

Question 2.

Evaluate the following :

(i) tan^{-1}(1) + cos^{-1}\(\left(\frac{1}{2}\right)\) + sin^{-1}\(\left(\frac{1}{2}\right)\)

Solution:

(ii) cos^{-1}\(\left(\frac{1}{2}\right)\) + 2 sin^{-1}\(\left(\frac{1}{2}\right)\)

Solution:

(iii) tan^{-1}\(\sqrt {3}\) – sec^{-1}(-2)

Solution:

∴ tan^{-1}\(\sqrt {3}\) – sec^{-1}(-2)

= \(\frac{\pi}{3}-\frac{2 \pi}{3}\) …[By (1) and (2)]

= –\(\frac{\pi}{3}\).

(iv) cosec^{-1}( \(-\sqrt{2}\)) + cot^{-1}(\(\sqrt{3}\))

Solution:

Question 3.

Prove the following :

(i) sin^{-1}\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^{-1}\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(-\frac{3 \pi}{4}\)

Question is modified.

sin^{-1}\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^{-1}\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(\frac{3 \pi}{4}\)

Solution:

(ii) sin^{-1}\(\left(-\frac{1}{2}\right)\) + cos^{-1}\(\left(-\frac{\sqrt{3}}{2}\right)\) = cos^{-1}\(\left(-\frac{1}{2}\right)\)

Solution:

(iii) sin^{-1}\(\left(\frac{3}{5}\right)\) + cos^{-1}\(\left(\frac{12}{13}\right)\) = sin^{-1}\(\left(\frac{56}{65}\right)\)

Solution:

(iv) cos^{-1}\(\left(\frac{3}{5}\right)\) + cos^{-1}\(\left(\frac{4}{5}\right)\) = \(\frac{\pi}{2}\)

Solution:

Let cos^{-1}\(\left(\frac{3}{5}\right)\) = x

∴ cosx = \(\left(\frac{3}{5}\right)\), where 0 < x < \(\frac{\pi}{2}\) ∴ sinx > 0

(v) tan^{-1}\(\left(\frac{1}{2}\right)\) + tan^{-1}\(\left(\frac{1}{3}\right)\) = \(\frac{\pi}{4}\)

Solution:

LHS = tan^{-1}\(\left(\frac{1}{2}\right)\) + tan^{-1}\(\left(\frac{1}{3}\right)\)

= tan^{-1}\(\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right)\)

= tan^{-1}\(\left(\frac{3+2}{6-1}\right)\) = tan^{-1}(1)

= tan^{-1}\(\left(\tan \frac{\pi}{4}\right)\) = \(\frac{\pi}{4}\)

= RHS.

(vi) 2 tan^{-1}\(\left(\frac{1}{3}\right)\) = tan^{-1}\(\left(\frac{3}{4}\right)\)

Solution:

(vii) tan^{-1}\(\left[\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right]\) = \(\frac{\pi}{4}\) + θ if θ ∈ \(\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\)

Solution:

(viii) tan^{-1}\(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\frac{\theta}{2}\), if θ ∈ (0, π)

Solution:

= \(\frac{\theta}{2}\) …[∵ tan^{-1}(tanθ) = θ]

= RHS.