These NCERT Solutions for Class 12 Maths Chapter 2 प्रतिलोम त्रिकोणमितीय फलन Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 2 प्रतिलोम त्रिकोणमितीय फलन Ex 2.2
प्रश्न 1.
3 sin-1 x = sin-1 (3x – 4x3), x ∈ \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
हल :
माना sin-1 x = θ
⇒ sin θ = x
R.H.S. = sin-1 (3x – 4x3)
= sin-1 (3 sin θ – 4 sin3 θ)
= sin-1(sin 3θ)
= 3θ = 3sin-1 x
R.H.S. = R.H.S.
अतः 3sin-1 x = sin-1 (3x – 4x3) इति सिद्धम् ।
दूसरी विधि:
माना sin x = θ
sin θ = x
∵ sin 3θ = 3 sin θ – 4 sin3 θ
⇒ sin 3θ = 3x – 4x3
⇒ 3θ = sin-1 (3x – 4x3)
अतः 3 sin-1 x = sin-1 (3x-4x3) इति सिद्धम् ।
प्रश्न 2.
3 cos-1 x = cos-1 (4x3 – 3x), x ∈ \(\left[\frac{1}{2}, 1\right]\)
हल:
माना cos-1 x = θ
⇒ cos θ = x
⇒ cos 3θ = 4 cos3 θ – 3 cos θ
⇒ cos 3θ = 4x3 – 3x
⇒ 3θ cos-1 (4x3-3x)
अतः 3 cos-1 x cos-1 (4x3 – 3x) इति सिद्धम् ।
प्रश्न 3.
tan-1 \(\frac{2}{11}\) + tan-1 \(\frac{7}{24}\) = tan-1 \(\frac{1}{2}\)
हल:
L.H.S. = tan-1 \(\frac{2}{11}\) + tan-1 \(\frac{7}{24}\)
प्रश्न 4.
2 tan-1 \(\frac{1}{2}\) + tan-1 \(\frac{1}{7}\) = tan-1 \(\frac{31}{17}\)
हल:
प्रश्न 5.
tan-1 \(\frac{1}{7}\) x ≠ 0
हल:
माना x=tan θ ⇒ θ = tan-1 x
प्रश्न 6.
tan-1 \(\frac{1}{\sqrt{x^2-1}}\) ,| x | > 1.
हल:
माना x=sec θ या θ =sec-1 x
प्रश्न 7.
tan-1 \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\) , x < π
हल:
प्रश्न 8.
tan-1 \(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\) , x < π
हल:
प्रश्न 9.
tan-1 \(\frac{x}{\sqrt{a^2-x^2}}\) ,| x | > a.
हल:
प्रश्न 10.
tan-1 \(\left\{\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right\}\), a > 0, – \(\frac{a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}\)
हल:
मान x = a tan θ या θ = tan-1 \(\frac {x}{a}\)
अब tan-1 \(\left\{\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right\}\)
= 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 \(\left\{\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right\}\)
= tan-1 (tan 3θ) [∴ tan 3x = \(\left(\frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}\right)\)]
= 3θ = 3 tan-1 \(\frac {x}{a}\)
अत: tan-1\(\left\{\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right\}\) का सरलतम रूप 3 tan-1 \(\frac {x}{a}\) है।
प्रश्न 11.
tan-1 \(\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)
हल:
tan-1 \(\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)
= tan-1 \(\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]\) (∵sin-1 \(\frac {1}{2}\) = \(\frac {π}{6}\)
= tan-1 \(\left[2 \cos \left(\frac{\pi}{3}\right)\right]\)
= tan-1 \(\left(2 \times \frac{1}{2}\right)\) (∵sin-1 \(\frac {π}{3}\) = \(\frac {1}{2}\)
= tan-1(1) = \(\frac {π}{4}\)
अत: tan-1 \(\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\) = \(\frac {π}{4}\)
प्रश्न 12.
cot (tan-1 a + cot-1 a)
हल:
cot (tan-1 a + cot-1 a)
= cot \(\frac {π}{2}\) (∵ tan-1 a + cot-1 a = \(\frac {π}{2}\) ) (∵cot \(\frac {π}{2}\) = 0)
अत: cot (tan-1 a + cot-1 a) = 0
प्रश्न 13.
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 तथा
हल:
tan \(\frac {1}{2}\) \(\left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right]\)
= tan [tan-1 x + tan-1 y] [∵2 tan-1 x = sin-1 \(\frac{2 x}{1+x^2}\) = cos-1\(\frac{1-x^2}{1+x^2}\)]
= tan [tan-1 \(\frac{x+y}{1-xy}\)]
= \(\frac{x+y}{1-xy}\)
अत: tan \(\frac{1}{2}\) \(\left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right]\) = \(\frac{x+y}{1-xy}\)
प्रश्न 14.
यदि sin \(\left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)\) = 1 तो x का मान ज्ञात कीजिए।
हल:
sin \(\left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)\) = 1
⇒ sin-1 + cos-1 x = sin-1 1
⇒ sin-1\(\frac{1}{5}\) + cos-1x = \(\frac{π}{2}\)
⇒ sin-1\(\frac{1}{5}\) = \(\frac{π}{2}\) – cos-1x
⇒ sin-1\(\frac{1}{5}\) = sin-1 x (∵sin-1 x + cos-1x = \(\frac{π}{2}\))
अत: x = \(\frac{1}{5}\)
प्रश्न 15.
यदि tan-1 \(\frac{x-1}{x-2}\) + tan-1\(\frac{x-1}{x-2}\) = \(\frac{π}{4}\),
तो x का मान ज्ञात कीजिए।
हल:
tan-1 \(\frac{x-1}{x-2}\) + tan-1\(\frac{x-1}{x-2}\) = \(\frac{π}{4}\)
⇒ \(\frac{2 x^2-4}{-3}=1\)
⇒ 2x2 – 4 = – 3
⇒ 2x2 = 4 – 3
⇒ 2x2 = 1
⇒ x2 = \(\frac{1}{2}\)
x = ± \(\frac{1}{\sqrt{2}}\)
प्रश्न संख्या 16 से 18 तक दिए प्रत्येक व्यंजक का मान ज्ञात कीजिए:
प्रश्न 16.
sin-1 sin \(\left(\sin \frac{2 \pi}{3}\right)\)
हल:
sin-1 की मुख्य मान शाखा \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) है।
∵ sin-1 \(\left(\sin \frac{2 \pi}{3}\right)\) = sin-1\(\left[\sin \left(\pi-\frac{\pi}{3}\right)\right]\)
= sin-1\(\left(\sin \frac{pi}{3}\right)\)
= \(\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
अत: sin-1\(\left(\sin \frac{2 \pi}{3}\right)=\frac{\pi}{3}\)
प्रश्न 17.
tan-1 \(\left(\tan \frac{3 \pi}{4}\right)\)
हल:
tan-1 की मुख्य मान शाखा \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) है।
∴ tan-1\(\left(\tan \frac{3 \pi}{4}\right)\) = tan-1\(\left[\tan \left(\pi-\frac{\pi}{4}\right)\right]\)
जहाँ \(\frac{3 \pi}{4} \notin\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
= tan-1 \(\left(-\tan \frac{\pi}{4}\right)\)
= tan-1 \(\left[\tan \left(-\frac{\pi}{4}\right)\right]\)
= – \(\frac{π}{2}\), जहाँ – \(\frac{\pi}{4} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
अत: tan-1\(\left(\tan \frac{3 \pi}{4}\right)\) = \(\frac{π}{4}\)
प्रश्न 18.
tan-1 \(\left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
हल:
माना sin-1 \(\frac{3}{5}\) = θ
∴ sin θ = \(\frac{3}{5}\)
तब cos θ = \(\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{1-\frac{9}{25}}\)
या cos θ = \(\frac{4}{5}\)
तब tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{\frac{3}{5}}{\frac{4}{5}}\)
∵ tan θ = \(\frac{3}{4}\) या θ = tan-1 \(\frac{3}{4}\)
अत: sin-1 \(\frac{3}{4}\) = tan-1 \(\frac{3}{4}\)
∴ tan \(\left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
= tan \(\left[\left(\tan ^{-1} \frac{3}{4}+\cot ^{-1} \frac{3}{2}\right)\right]\)
= tan \(\left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)\) (∵cot-1 x = tan-1\(\frac{3}{5}\))
= tan \(\left[\tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right]\)
= tan \(\left[\tan ^{-1} \frac{9+8}{12-6}\right]\) = tan \(\left[\tan ^{-1} \frac{17}{6}\right]=\frac{17}{6}\)
अत: tan \(\left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)=\frac{17}{6}\)
प्रश्न 19.
cos-1\(\left(\cos \frac{7 \pi}{6}\right)\) का मान बराबर है –
(A) \(\frac{7π}{6}\)
(B) \(\frac{5π}{6}\)
(C) \(\frac{π}{3}\)
(D) \(\frac{π}{6}\)
हल:
cos-1 की मुख्य मान शाखा [0, π] है।
∴ cos-1(cos \(\frac{7π}{6}\)) = cos-1[cos\(\left(2 \pi-\frac{5 \pi}{6}\right)\)]
= cos-1 \(\left[\cos \frac{5 \pi}{6}\right]\)
= \(\frac{5π}{6}\) जहाँ \(\frac{5π}{6}\) ∈ [0,π]
अतः विकल्प (B) सही है।
प्रश्न 20.
sin \(\left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]\) का मान है:
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{4}\)
(D) 1.
हल:
sin \(\left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]\)
= sin \(\left[\frac{\pi}{3}-\left(-\sin ^{-1} \frac{1}{2}\right)\right]\) [∴ sin-1(-x) = – sin-1 x]
= sin \(\left(\frac{\pi}{3}+\sin ^{-1} \frac{1}{2}\right)\)
= sin \(\left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin \left(\frac{\pi}{2}\right)\) = 1
अत: विकल्प (D) सही है।
प्रश्न 21.
tan-1 √3 – cot-1 (- √3) का मान है:
(Α) π
(B) – \(\frac{π}{2}\)
(C) 0
(D) \(2 \sqrt{3}\)
हल:
tan-1\(\sqrt{3}\) = tan-1 (tan \(\frac{π}{3}\)) = \(\frac{π}{3}\)
परन्तु cot-1 की मुख्य मान शाखा (0, π) है।
∴ cot-1 (-\(\sqrt{3}\)) = cot-1(- cot \(\frac{π}{6}\))
= cot-1 \(\left[\cot \left(\pi-\frac{\pi}{6}\right)\right]\)
= cot-1 \(\left[\cot \frac{5 \pi}{6}\right]=\frac{5 \pi}{6}\)
अब tan-1\(\sqrt{3}\) – cot-1(-\(\sqrt{3}\))
= \(\frac{\pi}{3}-\frac{5 \pi}{6}=\frac{2 \pi-5 \pi}{6}\)
= \(-\frac{3 \pi}{6}=-\frac{1}{2} \pi\)
= – \(\frac{π}{2}\)
अतः विकल्प (B) सही है।