Mathematics Multiple Choice Questions on “Inverse Trigonometric Functions Basics”.
1. (tan^{-1}sqrt{3}+sec^{-1}2 – cos^{-1}1) is equal to ________
a) 0
b) (frac{2π}{3})
c) (frac{π}{3})
d) (frac{π}{4})
Answer: b
Clarification: (tan^{-1}sqrt{3}=frac{π}{3}, sec^{-1}2=frac{π}{3}, cos^{-1}1=0)
∴(tan^{-1}sqrt{3}+sec^{-1}2 -cos^{-1}1=frac{π}{3}+frac{π}{3}-0)
=(frac{2π}{3}).
2. What is the principle value of (sec^{-1}(frac{2}{sqrt{3}})).
a) (frac{π}{6})
b) (frac{π}{3})
c) (frac{π}{4})
d) (frac{π}{2})
Answer: a
Clarification: Let (sec^{-1}(frac{2}{sqrt{3}}))=y
sec y=(frac{2}{sqrt{3}})
sec y=sec (frac{π}{6})
⇒y=(frac{π}{6})
3. What is the value of (tan^1frac{1}{√3}-sin^{-1}1+ cos^{-1}frac{1}{2}) is ________
a) 2π
b) (frac{π}{2})
c) π
d) 0
Answer: c
Clarification: (tan^{-1}frac{1}{sqrt{3}}=frac{π}{6},sin^{-1}1=frac{π}{2}, cos^{-1}frac{1}{2}=frac{π}{3})
(tan^1frac{1}{sqrt{3}}-sin^{-1}1+ cos^{-1}frac{1}{2}=frac{π}{6}+frac{π}{2}+frac{π}{3}=frac{π+3π+2π}{6}=frac{6π}{6}=π)
4. [-1, 1] is the domain for which of the following inverse trigonometric functions?
a) sin-1x
b) cot-1x
c) tan-1x
d) sec-1x
Answer: a
Clarification: [-1, 1] is the domain for sin-1x.
The domain for cot-1x is (-∞,∞).
The domain for tan-1x is (-∞,∞).
The domain for sec-1x is (-∞,-1]∪[1,∞).
5. The domain of sin-1(3x) is equal to _______
a) [-1, 1]
b) ([frac{-1}{3}, frac{1}{3}])
c) [-3, 3]
d) [-3π, 3π]
Answer: b
Clarification: The domain of y=sin-1x is -1≤x≤1.
∴the domain of y=sin-13x is-1≤3x≤1
⇒ ([frac{-1}{3} ≤ x ≤ frac{1}{3}])
Hence, ([frac{-1}{3}, frac{1}{3}]).
6. What is the value of 5 (cos^{-1}frac{1}{2} + 7 sin^{-1}(frac{-1}{2})) ?
a) –(frac{π}{2})
b) π
c) (frac{π}{2})
d) (frac{17π}{6})
Answer: c
Clarification: (cos^{-1}(frac{1}{2})=frac{π}{3} and sin^{-1}(-frac{1}{2})=-frac{π}{6})
∴ 5 (cos^{-1}frac{1}{2}+7 sin^{-1}(-frac{1}{2}) =5(frac{π}{3})+7(-frac{π}{6}))
=(frac{5π}{3}-frac{7π}{6}=frac{10π-7π}{6}=frac{3π}{6}=frac{π}{2})
7. Find the value of (sin^{-1}(sin frac{4π}{3})) is _______
a) π
b) (frac{π}{3})
c) (frac{4π}{3})
d) –(frac{π}{3})
Answer: d
Clarification: (sin^{-1}(sinx))=x, x∈([-frac{π}{2},frac{π}{2}])
∴(sin^{-1} (sin frac{4π}{3})=sin^{-1}(sin(π+frac{π}{3}))=sin^{-1}(sin(frac{-π}{3}))= -frac{π}{3}).
8. Find the value of (cos(sin^{-1}frac{sqrt{3}}{2})) is _____
a) (frac{sqrt{3}}{2})
b) (frac{1}{4})
c) (frac{1}{2})
d) 0
Answer: c
Clarification: (sinfrac{π}{3}=frac{sqrt{3}}{2})
∴(sin^{-1}frac{sqrt{3}}{2}=frac{π}{3})
⇒(cos(sin^{-1}frac{sqrt{3}}{2})=cos(frac{π}{3})=frac{1}{2}).
9. If (cos^{-1}x=y), then which of the following is correct?
a) 0 ≤ y ≤ π
b) 0 < y < π
c) –(frac{π}{2}≤y≤frac{π}{2})
d) –(frac{π}{2}
Clarification: Given that, (cos^{-1}x=y)
The range of principle values for the inverse trigonometric function (cos^{-1}) is [0,π].
Hence, 0≤y≤π.
10. (sin^{-1}x) is same as ((sinx)^{-1}).
a) True
b) False
Answer: b
Clarification: The given statement is false. (sin^{-1}x) is not same as ((sinx)^{-1}). (sin^{-1}x) is an inverse trigonometric function whereas ((sinx)^{-1}) is just the reciprocal of sinx i.e. (sinx=frac{1}{sinx}).