NCERT Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Class 12 Mathematics
Chapter-2 (Inverse Trigonometric Functions)
Questions and answers given in practice
Chapter-2 (Inverse Trigonometric Functions)
Exercise 2.1
Find
the principal values of the following:
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 [-π/2,π/2] and sin(-π/6)=-1/2.
Therefore, principal value of sin-1(-1/2)=-π/6.
Question 2. cos-1(√3/2)
Solution:
Let cos-1(√3/2) = y then, cos y = √3/2
Range of principal value
for cos-1 is [0, π] and cos(π/6) = √3/2
Therefore, principal value of cos-1(√3/2) = π/6.
Question
3. cosec-1(2)
Solution:
Let cosec-1(2) = y then, cosec y = 2
Range of principal value
for cosec-1 is [-π/2, π/2] -{0} and cosec(π/6) = 2
Therefore, principal value of cosec-1(2) = π/6.
Question
4: tan-1(-√3)
Solution:
Let tan-1(-√3) = y then, tan y = -√3
Range of principal value
for tan-1 is (-π/2, π/2) and tan(-π/3) = -√3
Therefore, principal value of tan-1(-√3) = -π/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, π] and cos(2π/3) = -1/2
Therefore, principal value of cos-1(-1/2) = 2π/3.
Question
6. tan-1(-1)
Solution:
Let tan-1(-1) = y then, tan y = -1
Range of principal value
for tan-1 is (-π/2, π/2) and tan(-π/4) = -1
Therefore, principal value of tan-1(-1) = -π/4.
Question
7. sec-1(2/√3)
Solution:
Let sec-1(2/√3) = y then, sec y = 2/√3
Range of principal value
for sec-1 is [0, π] – {π/2} and sec(π/6) = 2/√3
Therefore, principal value of sec-1(2/√3) = π/6.
Question
8. cot-1(√3)
Solution:
Let cot-1(√3) = y then, cot y = √3
Range of principal value
for cot-1 is (0, π) and cot(π/6) = √3
Therefore, principal value of cot-1(√3) = π/6.
Question
9. 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, π] and cos(2π/3) = -1/2
Therefore, principal value of cos-1(-1/2) = 3π/4.
Question
10. cosec-1(-√2)
Solution:
Let cosec-1(-√2) = y then, cosec y = -√2
Range of principal value
for cosec-1 is [-π/2, π/2] -{0} and cosec(-π/4) = -√2
Therefore, principal value of cosec-1(-√2) = -π/4.
Find
the values of the following:
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 [-π/2, π/2] and sin(-π/6) = -1/2.
Therefore, principal
value of sin-1(-1/2) = -π/6.
Let cos-1(-1/2) = x then, cos x = -1/2
Range of principal value
for cos-1 is [0, π] and cos(2π/3) = -1/2
Therefore, principal
value of cos-1(-1/2) = 2π/3.
Let tan-1(1) = z then, tan z = -1
Range of principal value
for tan-1 is (-π/2, π/2) and tan(π/4) = 1
Therefore, principal
value of tan-1(1) = π/4.
Now, tan-1(1) + cos-1(-1/2) + sin-1(-1/2) = π/4 + 2π/3 –
π/6
Adding them we will
get,
= (3π + 8π –
2π)/12
= 9π/12
= 3π/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 [-π/2, π/2] and sin(π/6) = 1/2.
Therefore, principal
value of sin-1(1/2) = π/6.
Let cos-1(1/2) = x then, cos x = 1/2
Range of principal value
for cos-1 is [0, π] and cos(π/3) = 1/2
Therefore, principal
value of cos-1(1/2) = π/3.
Now, cos-1(1/2) + 2 sin-1(1/2) = π/3 + 2π/6
Adding them we will get,
= (2π + 2π)/6
= 4π/6
= 2π/3
Question
13. If sin–1 x = y, then
(A) 0 ≤ y ≤ π (B) -π / 2 ≤y ≤ π / 2 (C) 0 < y
< π (D) -π / 2 <y < π / 2
Solution:
We know that the
principal range for sin-1 is [-π / 2, π / 2]
Hence, if sin-1 x = y, y
€ [-π / 2, π / 2]
Therefore, -π / 2 ≤y ≤ π
/ 2.
Hence, option (B) is correct.
Question
14. tan–1(√3) – sec-1(-2)
is equal to
(A)
π (B) -π/3 (C) π/3 (D) 2π/3
Solution:
For solving this
question we will use principal values of sec-1 & tan-1
Let tan-1(√3) = y then, tan y = √3
Range of principal value
for tan-1 is (-π/2, π/2) and tan(π/3) = √3
Therefore, principal
value of tan-1(√3) = π/3.
Let sec-1(-2) = y then, sec y = -2
Range of principal value
for sec-1 is [0, π] – {π/2} and sec(2π/3) = – 2
Therefore, principal
value of sec-1(-2) = 2π/3.
Now, tan–1 (√3) – sec -1(-2)
= π/3 – 2π/3
= -π/3
Hence, option (B) is correct.
