Practice Question On Inverse Trigonometry For IITJEE , NDA and Airforce 16-30 (Set 2 )

Inverse Trigonometry

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Question no. 16

what is equal to ?

looks_one 0

looks_two 2

looks_3 1

looks_4 1/2

Option looks_3 1

Solution :

Question no. 17

If sin^-1(1) +sin^-1(4/5)=sin^-1(x) then what is x equal to ?

looks_one 3/5

looks_two 4/5

looks_3 1

looks_4 0

Option looks_one 3/5

Solution :

Question no. 18

tan^-1(1+x) +tan^-1(1-x)=π/2 is satisfied by

looks_one x=1

looks_two x=-1

looks_3 x=0

looks_4 x=1/2

Option looks_3 x=0

Solution :

Question no. 19

If , where x=

looks_one 1/√2

looks_two -1/√2

looks_3 ∓ √(5/2)

looks_4 ∓ 1/2

Option looks_4 ∓ 1/2

Solution :

Question no. 20

tan^-11 + tan^-12 + tan^-13=

looks_one π/2

looks_two π

looks_3 π/4

looks_4 none of these

Option looks_two π

Solution :
⇒ tan^-11 + tan^-12 + tan^-13
⇒ tan^-11 + + tan^-13 + tan^-12
⇒ tan^-1(1+3)/(1-3) + tan^-12
⇒ tan^-1(-2) + tan^-12
⇒ π -tan^-12 + tan^-12 = π

Question no. 21

cos^-1x + cos^-1y + cos^-1z= π , then

looks_one x²+y² + z² +xyz = 0

looks_two x²+y² + z² +2xyz = 0

looks_3 x²+y² + z² +xyz = 1

looks_4 x²+y² + z² +2xyz = 1

Option looks_4 x²+y² + z² +2xyz = 1

Solution :

Question no. 22

If 2tan^-1(1/3) + tan^-1(1/2) =

looks_one 90^o

looks_two 60^o

looks_3 45^o

looks_4 tan^-12

Option looks_3 45^o

Solution :

Question no. 23

sin^-1x + sin^-1(1/x) + cos^-1x +cos^-1(1/x)=

looks_one π

looks_two π/2

looks_3 3π/2

looks_4 none of these

Option looks_one π

Solution :
⇒ sin^-1x + sin^-1(1/x) + cos^-1x +cos^-1(1/x)
⇒ (sin^-1x + cos^-1x) + (sin^-1(1/x) + cos^-1(1/x))
⇒ π/2 + π/2 = π

Question no. 24

tan^-1x + tan^-1y + tan^-1z =π ,then

looks_one x+y+z -xyz=0

looks_two x+y+z +xyz=0

looks_3 xy+yz+xz+1=0

looks_4 xy+yz+xz-1=0

Option looks_one x+y+z -xyz=0

Solution :

Question no. 25

If sin(cot^-1(x+1))=cos(tan^-1x), then x=

looks_one -1/2

looks_two 1/2

looks_3 0

looks_4 9/4

Option looks_one -1/2

Solution :

Question no. 26

If u= cot^-1√tanθ -tan^-1√tanθ , then =

looks_one √tanθ

looks_two √cotθ

looks_3 cotθ

looks_4 tanθ

Option looks_one √tanθ

Solution :
u= cot^-1√tanθ -tan^-1√tanθ

Question no. 27

If tan^-1(1/4) +tan^-1(2/9) is equal

looks_one 0.5cos^-1(3/5)

looks_two 0.5sin^-1(3/5)

looks_3 0.5tan^-1(3/5)

looks_4 tan^-1(1/2)

Option looks_4 tan^-1(1/2)

Solution :

Question no. 28

If cot^-1√(cosβ) -tan^-1√(cosβ)= x , then the value of sinx is

looks_one tan²(β/2)

looks_two cot²(β/2)

looks_3 tanβ

looks_4 cot(β /2)

Option looks_one tan²(β/2)

Solution :
cot^-1√(cosβ) -tan^-1√(cosβ)= x

Question no. 29

The trigonometric equation sin^-1x = 2sin^-1a has a solution for

looks_one all values of a

looks_two |a| 1/√2

looks_3|a| ≥ 1/√2

looks_4 -1/√2< a < 1/√2

Option looks_4 -1/√2< a < 1/√2

Solution :

Question no. 30

If cos^-1x -cos^-1(y/2)=α , then 4x² -4xycosα +y² is

looks_one 4

looks_two 4sin²α

looks_3 -4sin²α

looks_4 2sin²α

Option looks_two 4sin²α

Solution :

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