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Class 12 NCERT Solutions – Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.1

• Difficulty Level : Medium
• Last Updated : 25 Feb, 2021

Question 1. sin-1(-1/2)

Solution:

Let sin-1(-1/2) = y then, sin y = -1/2

Range of principal value for sin-1 is [-Ï€/2,Ï€/2] and sin(-Ï€/6)=-1/2.

Therefore, principal value of sin-1(-1/2)=-Ï€/6.

Question 2. cos-1(âˆš3/2)

Solution:

Let cos-1(âˆš3/2) = y then, cos y = âˆš3/2

Range of principal value for cos-1 is [0, Ï€] and cos(Ï€/6) = âˆš3/2

Therefore, principal value of cos-1(âˆš3/2) = Ï€/6.

Question 3. cosec-1(2)

Solution:

Let cosec-1(2) = y then, cosec y = 2

Range of principal value for cosec-1 is [-Ï€/2, Ï€/2] -{0}  and cosec(Ï€/6) = 2

Therefore, principal value of cosec-1(2) = Ï€/6.

Question 4: tan-1(-âˆš3)

Solution:

Let tan-1(-âˆš3) = y then, tan y = -âˆš3

Range of principal value for tan-1 is (-Ï€/2, Ï€/2) and tan(-Ï€/3) = -âˆš3

Therefore, principal value of tan-1(-âˆš3) = -Ï€/3.

Question 5. cos-1(-1/2)

Solution:

Let cos-1(-1/2) = y then, cos y = -1/2

Range of principal value for cos-1 is [0, Ï€] and cos(2Ï€/3) = -1/2

Therefore, principal value of cos-1(-1/2) = 2Ï€/3.

Question 6. tan-1(-1)

Solution:

Let tan-1(-1) = y then, tan y = -1

Range of principal value for tan-1 is (-Ï€/2, Ï€/2)  and tan(-Ï€/4) = -1

Therefore, principal value of tan-1(-1) = -Ï€/4.

Question 7. sec-1(2/âˆš3)

Solution:

Let sec-1(2/âˆš3) = y then, sec y = 2/âˆš3

Range of principal value for sec-1 is [0, Ï€] – {Ï€/2} and sec(Ï€/6) = 2/âˆš3

Therefore, principal value of sec-1(2/âˆš3) = Ï€/6.

Question 8. cot-1(âˆš3)

Solution:

Let cot-1(âˆš3) = y then, cot y = âˆš3

Range of principal value for cot-1 is (0, Ï€) and cot(Ï€/6) = âˆš3

Therefore, principal value of cot-1(âˆš3) = Ï€/6.

Question 9. cos-1(-1/âˆš2)

Solution:

Let cos-1(-1/âˆš2) = y then, cos y = -1/âˆš2

Range of principal value for cos-1 is [0, Ï€] and cos(2Ï€/3) = -1/2

Therefore, principal value of cos-1(-1/2) = 3Ï€/4.

Question 10. cosec-1(-âˆš2)

Solution:

Let cosec-1(-âˆš2) = y then, cosec y = -âˆš2

Range of principal value for cosec-1 is [-Ï€/2, Ï€/2] -{0} and cosec(-Ï€/4) = -âˆš2

Therefore, principal value of cosec-1(-âˆš2) = -Ï€/4.

Question 11. tan-1(1) + cos-1(-1/2) + sin-1(-1/2)

Solution:

For solving this question we will use principal values of sin-1, cos-1 & tan-1

Let sin-1(-1/2) = y then, sin y = -1/2

Range of principal value for sin-1 is [-Ï€/2, Ï€/2] and sin(-Ï€/6) = -1/2.

Therefore, principal value of sin-1(-1/2) = -Ï€/6.

Let cos-1(-1/2) = x then, cos x = -1/2

Range of principal value for cos-1 is [0, Ï€] and cos(2Ï€/3) = -1/2

Therefore, principal value of cos-1(-1/2) = 2Ï€/3.

Let tan-1(1) = z then, tan z = -1

Range of principal value for tan-1 is (-Ï€/2, Ï€/2)  and tan(Ï€/4) = 1

Therefore, principal value of tan-1(1) = Ï€/4.

Now, tan-1(1) + cos-1(-1/2) + sin-1(-1/2) = Ï€/4 + 2Ï€/3 – Ï€/6

= (3Ï€ + 8Ï€ – 2Ï€)/12

= 9Ï€/12

= 3Ï€/4

Question 12. cos-1(1/2) + 2 sin-1(1/2)

Solution:

For solving this question we will use principal values of sin-1 & cos-1

Let sin-1(1/2) = y then, sin y = -1/2

Range of principal value for sin-1 is [-Ï€/2, Ï€/2] and sin(Ï€/6) = 1/2.

Therefore, principal value of sin-1(1/2) = Ï€/6.

Let cos-1(1/2) = x then, cos x = 1/2

Range of principal value for cos-1 is [0, Ï€] and cos(Ï€/3) = 1/2

Therefore, principal value of cos-1(1/2) = Ï€/3.

Now, cos-1(1/2) + 2 sin-1(1/2) = Ï€/3 + 2Ï€/6

= (2Ï€ + 2Ï€)/6

= 4Ï€/6

= 2Ï€/3

(A) 0 â‰¤ y â‰¤ Ï€ (B) -Ï€ / 2 â‰¤y â‰¤ Ï€ / 2 (C) 0 < y < Ï€ (D) -Ï€ / 2 <y < Ï€ / 2

Solution:

We know that the principal range for sin-1 is [-Ï€ / 2, Ï€ / 2]

Hence, if sin-1 x = y, y â‚¬ [-Ï€ / 2, Ï€ / 2]

Therefore, -Ï€ / 2 â‰¤y â‰¤ Ï€ / 2.

Hence, option (B) is correct.

(A) Ï€ (B) -Ï€/3 (C) Ï€/3 (D) 2Ï€/3

Solution:

For solving this question we will use principal values of sec-1 & tan-1

Let tan-1(âˆš3) = y then, tan y = âˆš3

Range of principal value for tan-1 is (-Ï€/2, Ï€/2) and tan(Ï€/3) = âˆš3

Therefore, principal value of tan-1(âˆš3) = Ï€/3.

Let sec-1(-2) = y then, sec y = -2

Range of principal value for sec-1 is [0, Ï€] – {Ï€/2} and sec(2Ï€/3) = – 2

Therefore, principal value of sec-1(-2) = 2Ï€/3.

Now, tanâ€“1 (âˆš3) – sec -1(-2)

= Ï€/3 – 2Ï€/3

= -Ï€/3

Hence, option (B) is correct.

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