Practice question on Inverse Trigonometry for IIT-JEE , Airforce and NDA (Set 1 )

Inverse Trigonometry

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Question no. 1

The Value of 1+ cot²(sin^-1x)

looks_one 1/2x

looks_two 2/x

looks_3 x²

looks_4 1/x²

Option looks_4 1/x²

Solution :
1+ cot²(sin^-1x) = cosec²(sin^-1x)
= [cosec(cosec^-1(1/x))]² = [1/x]²
= 1/x²

Question no. 2

=

looks_one tan^-1x

looks_two

looks_3 2tan^-1x

looks_4 none of these

Option looks_two 1/2(tan^-1x)

Solution :

Question no. 3

cos(tan^-1x)=

looks_one

looks_two

looks_3 1+x²

looks_4 none of these

Option looks_two

Solution :

Question no. 4

=

looks_one

looks_two

looks_3

looks_4

Option

Solution :

Question no. 5

=

looks_one 1/x

looks_two x

looks_3

looks_4

Option looks_two x

Solution :

Question no. 6

sec²(tan^-12)+cosec²(cot^-13)=

looks_one 5

looks_two 13

looks_3 15

looks_4 6

Option looks_3 15

Solution :
sec²(tan^-12)+cosec²(cot^-13)
sec²(sec^-1√5)+cosec²(cosec^-1√10) = (√5)² + (√10)² = 15

Question no. 7

If , then x=

looks_one 1

looks_two √3

looks_3 1/√3

looks_4 none of these

Option looks_3 1/√3

Solution :

Question no. 8

If sin^-1x=θ + β and sin^-1= θ - β , then 1+xy=

looks_one sin²θ + sin²β

looks_two sin²θ + cos²β

looks_3 cos²θ + cos²β

looks_4 cos²θ + sin²β

Option looks_two sin²θ + cos²β

Solution :
sin^-1x=θ + β ⇒ x = sin ( θ + β )
sin^-1= θ - β ⇒ y = sin ( θ - β )
xy = sin ( θ + β ) . sin ( θ - β )
xy = sin²θ - sin²β
1 + xy = 1 + sin²θ - sin²β
1 + xy = sin²θ + cos²β

Question no. 9

cot[ cos^-1(7/25)]

looks_one 25/24

looks_two 25/7

looks_3 24/25

looks_4 none of these

Option looks_4 none of these

Solution :
cot[ cos^-1(7/25)]
cot[ cot^-1(7/24)]
7/24

Question no. 10

[sin(tan^-13/4)]² =

looks_one 3/5

looks_two 5/3

looks_3 9/25

looks_4 25/9

Option looks_3 9/25

Solution :
[sin(tan^-13/4)]²
[sin(sin^-13/5)]²
(3/5)² = 9 / 25

Question no. 11

The principal value of sin^-1(sin5π/3)

looks_one 5π/3

looks_two -5π/3

looks_3 -π/3

looks_4 4π/3

Option looks_3 -π/3

Solution :
Range of sinx is [- π/2 , π/2 ]
sin^-1(sin5π/3)
sin^-1(sin(2π - π/3 )
sin^-1(sin(- π/3)
sin^-1(-sin( π/3)) = -π/3

Question no. 12

The value of x which satisfies the equation tan^-1x = sin^-1(3/√10)

looks_one 3

looks_two -3

looks_3 1/3

looks_4 -1/3

Option looks_one 3

Solution :
tan^-1x = sin^-1(3/√10)
tan^-1x = tan^-13
x = 3

Question no. 13

sin[cot^-1(cos(tan^-1x))]=

looks_one

looks_two

looks_3

looks_4

Option looks_4

Solution :

Question no. 14

If sin^-1x +sin^-1y + sin^-1z = π/2, then the value of x² + y² + z² +2xyz is equal to

looks_one 0

looks_two 2

looks_3 3

looks_4 1

Option looks_one 0

Solution :

Question no. 15

If θ= sin^-1[sin(-600^o)], then one of the possible value of θ is

looks_one π/3

looks_two π/2

looks_3 2π/3

looks_4 -2π/3

Option looks_one π/3

Solution :
θ = sin^-1[-sin(540^o + 60^o)]
= sin^-1[-(-sin( 60^o))]
= sin^-1(60^o) = 60^o

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