GSEB Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1

Gujarat Board Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1

Find the principal values of the following:

Question 1.
sin^-1(-(frac { 1 }{ 2 }))
Solution:
Let sin^-1(-(frac { 1 }{ 2 })) = y
∴ sin y = (frac { 1 }{ 2 }) = – sin(frac { π }{ 6 }) = sin(-(frac { π }{ 6 }))
The range of principal value branch of sin^-1 is [-(frac { π }{ 2 }), (frac { π }{ 2 })].
Hence, principal value of sin^-1(-(frac { 1 }{ 2 })) is – (frac { π }{ 6 }).

Question 2.
cos^-1(left(-frac{sqrt{3}}{2}right))
Solution:
Let cos^-1(left(frac{sqrt{3}}{2}right)) = y
∴ cos y = (left(frac{sqrt{3}}{2}right)) = – cos(frac { π }{ 6 })
The range of principal value branch is (0, π)
∴ principal value of cos^-1(left(frac{sqrt{3}}{2}right)) is (frac { π }{ 6 }).

Question 3.
cosec^-1(2)
Solution:
Let cosec^-1(2) = y
∴ cosec y = 2 = cosec(frac { π }{ 6 })
The range of principal value branch of cosec^-1 is [-(frac { π }{ 2 }), (frac { π }{ 2 })] – [0]
∴ principal value of cosec^-1(2) is (frac { π }{ 6 }).

Question 4.
tan^-1(-(sqrt{3}))
Solution:
Let tan^-1(-(sqrt{3})) = y
∴ tan y = 2 = – (sqrt{3}) = – tan(frac { π }{ 3 }) = tan(-(frac { π }{ 3 }))
The range of principal value branch of tan^-1 is [-(frac { π }{ 2 }), (frac { π }{ 2 })]
∴ principal value of
tan^-1(-(sqrt{3})) is – (frac { π }{ 3 }).

Question 5.
cos^-1(-(frac { 1 }{ 2 }))
Solution:
Let cos^-1(-(frac { 1 }{ 2 })) = y
∴ cos y = –(frac { 1 }{ 2 }) = – cos(frac { π }{ 3 }) = cos(π – (frac { π }{ 3 })
= cos (frac { 2π }{ 3 })
The range of principal value branch is (0, π)
∴ principal value of cos^-1(-(frac { 1 }{ 2 })) is (frac { 2π }{ 3 }).

Question 6.
tan^-1(-1)
Solution:
Let tan^-1(-1) = y.
∴ tan y = – 1 = tan(frac { π }{ 4 })) = tan((frac { -π }{ 4 }))
The range of principal value branch of tan^-1 is [-(frac { π }{ 2 }), (frac { π }{ 2 })]
∴ principal value of tan^-1(-1) is (frac { π }{4 }).

Question 7.
sec^-1(left(-frac{2}{sqrt{3}}right))
Solution:
Let sec^-1(left(frac{2}{sqrt{3}}right)) = y
∴ sec y = (left(frac{2}{sqrt{3}}right)) = sec((frac { π }{ 6 }))
The range of principal value branch of sec^-1 is [0, π], ((frac { π }{ 2 }))
∴ principal value of sec^-1(left(frac{2}{sqrt{3}}right)) is (frac { π }{6 }).

Question 8.
cot^-1((sqrt{3}))
Solution:
Let cot^-1((sqrt{3})) = y
∴ cot y = (sqrt{3}) = cot(frac { π }{ 6 })
The range of principal value branch of cot^-1 is [0, π]
∴ principal value of cot^-1(sqrt{3}) is (frac { π }{ 6 }).

Question 9.
cos^-1(left(-frac{1}{sqrt{2}}right))
Solution:
Let cos^-1(left(-frac{1}{sqrt{2}}right)) = y
∴ cos y = (left(-frac{1}{sqrt{2}}right)) = cos((frac { π }{ 4 })) = cos(π – (frac { π }{ 4 }))
The range of principal value branch of cos^-1 is [0, π].
∴ principal value of cos^-1(left(-frac{1}{sqrt{2}}right)) is (frac { 3π }{4 }).

Question 10.
cosec^-1 (-(sqrt{2}))
Solution:
Let cosec^-1(-(sqrt{2})) = y
∴ cosec y = –(sqrt{2}) = – cosec(frac { π }{ 4 }) = cosec (- (frac { π }{ 4 })).
The range of principal value branch of cosec^-1 is (- (frac { π }{ 2 }), (frac { π }{ 2 })) – {0}
∴ principal value of cosec^-1(-(sqrt{2})) is – (frac { π }{ 4 }).

Question 11.
tan^-1(1) + cos^-1(-(frac { 1 }{ 2 })) + sin^-1((frac { -1 }{ 2 }))
Solution:
tan^-1(1) + cos^-1(-(frac { 1 }{ 2 })) + sin^-1((frac { -1 }{ 2 }))
Now, tan^-1 1 = (frac { π }{ 4 }),
Since range of principal value branch of cos^-1 is [0, π]. So,

Question 12.
cos^-1((frac { 1 }{ 2 })) + 2 sin^-1((frac { 1 }{ 2 }))
Solution:
cos^-1((frac { 1 }{ 2 })) + 2 sin^-1((frac { 1 }{ 2 }))
Now, cos^-1((frac { 1 }{ 2 })) = (frac { π }{ 3 })
and sin^-1((frac { 1 }{ 2 })) = (frac { π }{ 6 })
∴ cos^-1((frac { 1 }{ 2 })) + 2 sin^-1((frac { 1 }{ 2 }))
= (frac { π }{ 3 }) + 2 x (frac { π }{ 6 }) = (frac { 2π }{ 3 }).

Question 13.
If sin^-1 = y, then
(A) 0 ≤ y ≤ π
(B) – (frac { π }{ 2 }) ≤ y ≤ (frac { π }{ 2 })
(C) 0 < y < π
(D) – (frac { π }{ 2 }) < y (frac { π }{ 2 })
Solution:
The range of principal value branch of sin^-1 is [-(frac { π }{ 2 }), (frac { π }{ 2 })]
∴ if sin^-1 x = y, then – (frac { π }{ 2 }) ≤ y ≤ (frac { π }{ 2 })
∴ Part (B) is correct.

Question 14.
tan^-1(sqrt{3}) – sec^-1(-2) is equal to
(A) π
(B) – (frac { π }{ 3 })
(C) (frac { π }{ 3 })
(D) (frac { 2π }{ 3 })
Solution:
tan^-1(sqrt{3}) = (frac { π }{ 3 })
and sec^-1(-2) = π – (frac { π }{ 3 }) = (frac { 2π }{ 3 }),
since principal value branch of sec^-1 is [0, π] – {(frac { π }{ 2 })}
∴ tan^-1(sqrt{3}) – sec^-1(-2) = (frac { π }{ 3 }) – (frac { 2π }{ 3 })
= – (frac { π }{ 3 })
∴ Part (B) is correct.