Referring to the AP Inter 2nd Year Maths Study Material Chapter 2 Inverse Trigonometric Functions Exercise 2b Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Inverse Trigonometric Functions Solutions Exercise 2b
I.
Question 1.
Prove that 3sin-1x = sin-1(3x – 4x3), x ∈ \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
Solution:
Put x = sinθ ⇒ θ = sin-1x
Now RHS = sin-1(3x – 4x3) = sin-1(3sinθ – 4sin3θ) = sin-1(sin 3θ)
= 3θ = 3 sin-1x = LHS
Question 2.
Prove that 3cos-1x = cos-1 (4x3 – 3x), x ∈ [\(\frac{1}{2}\), 1]
Solution:
Put x = cosθ ⇒ θ = cos-1x
Now RHS = cos-1 (4x3 = 3x) = cos-1(4cos3θ – 3cosθ)
= cos-1(cos 3θ) = 3θ = 3 cos-1x = LHS [∵ cos-1(cosθ) = θ]
Question 3.
Find the value of sin-1 \(\left(\sin \frac{2 \pi}{3}\right)\).
Solution:
Here 2π/3 = 120° is not in the P.V range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) of sin-1x
∴ \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right]=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)=\frac{\pi}{3} \text { and } \frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Question 4.
Find the value of tan-1 \(\left(\tan \frac{3 \pi}{4}\right)\).
Solution:
Here 3π/4 = 135° is not in the P.V range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) of sin-1x
∴ \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{4}\right)\right]=\tan ^{-1}\left[-\tan \left(\frac{\pi}{4}\right)\right]\)
= \(-\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \text { and }-\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
II.
Question 1.
Find the value of tan(sin-1\(\frac{3}{5}\) + cot-1\(\frac{3}{2}\))
Solution:
Put sin-1\(\frac{3}{5}\) = θ ⇒ sin θ = \(\frac{3}{5}\). Hence tan θ = \(\frac{3}{4}\).
∴ sin-1\(\frac{3}{5}\) = tan-1\(\frac{3}{4}\) …………… (1)
cot-1\(\frac{3}{2}\) = tan-1\(\frac{2}{3}\) …………… (2)
∴ tan (sin-1\(\frac{3}{5}\) + cot-1\(\frac{3}{2}\)) = tan(tan-1\(\frac{3}{4}\) + tan-1\(\frac{2}{3}\))
= \(\tan \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right]=\tan \left(\tan ^{-1} \frac{17}{6}\right)=\frac{17}{6}\) [∵ tan-1x + tan-1y = \(\tan ^{-1}\left(\frac{x+y}{1-x y}\right) .\)]
Question 2.
Write the function tan-1\(\), x ≠ 0 in the simplest form.
Solution:
Put x = tanθ ⇒ θ = tan-1x
\(\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)\)
= \(\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x\) [∵ 1 – cos 2θ = 2 sin2 θ; sin 2θ = 2 cos θ sin θ]
Question 3.
Write the function tan-1\(\left(\sqrt{\frac{1-\cos \mathrm{x}}{1+\cos \mathrm{x}}}\right)\), 0 < x < π in the simplest form
Solution:
We know that 1 – cosx = 2 sin2\(\frac{x}{2}\) and 1 + cosx = 2 cos2\(\frac{x}{2}\). Also tan-1(tan θ) = θ
∴ \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\tan ^{-1}\left(\sqrt{\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}}\right)=\tan ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}\)
Question 4.
Write the function tan-1\(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\), \(\frac{-\pi}{4}\) < x < \(\frac{3\pi}{4}\) in the simplest form
Solution:
G.E = \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\tan ^{-1}\left(\frac{\frac{\cos x-\sin x}{\cos x}}{\frac{\cos x+\sin x}{}}\right)=\tan ^{-1}\left(\frac{\frac{\cos x}{\sin x}}{\cos x}\right)\)
= \(\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)\) = tan-1 (1) – tan-1 (tan x) = \(\frac{\pi}{4}\) – x [∵ tan-1\(\left(\frac{1-x}{1+x}\right)\) = tan-1 1 – tan-1 2]
Question 5.
Write the function tan-1\(\frac{x}{\sqrt{a^2-x^2}}\), |x| < a in the simplest form.
Solution:
Put x = a sin θ ⇒ sin θ = \(\frac{\mathrm{x}}{\mathrm{a}}\) ⇒ θ = sin-1 \(\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\)
∴ \(\tan ^{-1} \frac{x}{\sqrt{a^2-x^2}}=\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right)=\tan ^{-1}\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^2 \theta}}\right)\)
= \(\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)=\tan ^{-1}(\tan \theta)=\theta=\sin ^{-1}\left(\frac{x}{a}\right)\)
Question 6.
Write the function tan-1\(\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)\), a > 0; \(\frac{-a}{\sqrt{3}}\) < x < \(\frac{a}{\sqrt{3}}\) in the simplest form.
Solution:
Put x = a tan θ ⇒ tan θ = \(\frac{\mathrm{x}}{\mathrm{a}}\) ⇒ θ = tan-1 \(\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\)
∴ \(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=\tan ^{-1}\left(\frac{3 a^2 \cdot(a \tan \theta)-a^3 \tan ^3 \theta}{a^3-3 a \cdot\left(a^2 \tan ^2 \theta\right)}\right)\)
= \(\tan ^{-1}\left(\frac{3 a^3 \tan \theta-a^3 \tan ^3 \theta}{a^3-3 a^3 \tan ^2 \theta}\right)=\tan ^{-1}(\tan 3 \theta)=3 \theta=3 \tan ^{-1}\left(\frac{x}{a}\right)\)
Question 7.
Find the value of tan-1\(\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)
Solution:
\(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]=\tan ^{-1}\left[2 \cos \left(2 \cdot \frac{\pi}{6}\right)\right]=\tan ^{-1}\left[2\left(\frac{1}{2}\right)\right]=\tan ^{-1} 1=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}\)
Question 8.
Find the value of tan \(\frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right]\), |x| < 1, y > 0 and xy < 1
Solution:
Put x = tan θ ⇒ θ = tan-1 x
Now sin-1\(\frac{2 x}{+x^2}\) = sin-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = sin-1 (sin 2θ) = 2θ = 2 tan-1 x
Put y = tan Φ ⇒ Φ = tan-1 y
Now \(\cos ^{-1} \frac{1-y^2}{1+y^2}=\cos ^{-1}\left(\frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right)=\cos ^{-1}(\cos 2 \phi)=2 \phi=2 \tan ^{-1} y\)
∴ \(\tan \frac{1}{2}\left(\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right)=\tan \frac{1}{2}\left(2 \tan ^{-1} x+2 \tan ^{-1} y\right)\)
= \(\tan \left(\tan ^{-1} x+\tan ^{-1} y\right)=\tan \left[\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]=\frac{x+y}{1-x y}\)