inverse trigonometric functions

INVERSE TRIGONOMETRIC FUNCTIONS
PREVIOUS EAMCET BITS
⎛ −1 ⎞
⎛ −1 ⎞
⎛1⎞
cos −1 ⎜ ⎟ − 2sin −1 ⎜ ⎟ + 3cos −1 ⎜
− 4 tan −1 ( −1) =
⎟
⎝ 2 ⎠
⎝2⎠
⎝ 2⎠
19π
35π
47π
1)
2)
3)
12
12
12
Ans: 4
2π
π
3π
π 43π
Sol.
− 2 × + 3× + 4 × =
3
6
4
4 12
⎛3⎞
⎛4⎞ π
2.
If sin −1 ⎜ ⎟ + sin −1 ⎜ ⎟ = then x =
⎝x⎠
⎝x⎠ 2
1) 3
2) 5
3) 7
Ans: 2
3
4
⎛3⎞
⎛4⎞ π
Sol. sin −1 ⎜ ⎟ + sin −1 ⎜ ⎟ = ⇒ sin −1 = cos −1
x
x
⎝x⎠
⎝x⎠ 2
1.
[EAMCET 2009]
4)
43π
12
[EAMCET 2008]
4) 11
3
16
9
16
= sin −1 1 − 2 ⇒ 2 = 1 − 2 ⇒ x 2 = 25
x
x
x
x
⎛
⎛ 1 ⎞⎞
The value of x where x > 0 and tan ⎜ sec −1 ⎜ ⎟ ⎟ = sin ( tan −1 2 ) is
⎝ x ⎠⎠
⎝
⇒ sin −1
3.
1)
5
2)
5
3
3) 1
[EAMCET 2007]
4)
2
3
Ans: 2
1⎞
⎛
Sol. tan ⎜ sec −1 ⎟ = sin ( tan −1 2 )
x⎠
⎝
⎛
⎛
1 − x2 ⎞
2 ⎞
tan ⎜ tan −1
⎟ = sin ⎜ sin −1
⎟
⎜
x ⎟⎠
5⎠
⎝
⎝
1 − x2
2
5
⇒
=
⇒x=
x
3
5
4
1
sin −1 + 2 tan −1 =
4.
5
3
π
π
π
1)
2)
3)
3
4
2
Ans: 3
4
1
4
3
Sol. sin −1 + 2 tan −1 = sin −1 + tan −1
5
3
5
4
4
4 π
= sin −1 + cos −1 =
5
5 2
−1
−1
−1
5.
sin x + sin (1 − x ) = cos x ⇒ x ∈
1
[EAMCET 2005]
4) 0
[EAMCET 2004]
Inverse Trigonometric Functions
1) {1, 0}
2) {−1,1}
⎧ 1⎫
3) ⎨0, ⎬
⎩ 2⎭
4) {2, 0}
Ans: 3
π
− 2sin −1 ( x )
2
1
1 − x = 1 − 2x 2 ⇒ x = 0,
2
⎡
⎛1⎞
⎛ 1 ⎞⎤
6.
cos ⎢cos −1 ⎜ ⎟ + sin −1 ⎜ − ⎟ ⎥ = ……..
⎝7⎠
⎝ 7 ⎠⎦
⎣
1
2) 0
1) −
3
Ans: 2
⎡
⎛ 1⎞
⎛ 1 ⎞⎤
Sol. cos ⎢cos −1 ⎜ − ⎟ + sin −1 ⎜ − ⎟ ⎥
⎝ 7⎠
⎝ 7 ⎠⎦
⎣
Sol. sin −1 (1 − x ) =
π⎤
π
⎡
= 0 ⎢∵ sin −1 x + cos −1 x = ⎥
2
2⎦
⎣
π
7.
If sin −1 x − cos −1 x = , then x =
6
1
3
2)
1)
2
2
Ans: 2
Sol. By verification x = 3 / 2
8.
sec 2 ( tan −1 2 ) + cos ec 2 ( cot −1 3) =
[EAMCET 2003]
3)
1
3
4)
4
9
= cos
1) 5
2) 10
Ans: 3
Sol. Let tan −1 ( 2 ) = α and cot −1 ( 3) = β
[EAMCET 2002]
3)
−1
2
4)
− 3
2
[EAMCET 2001]
3) 15
4) 20
3) 5/14
4) 14/5
tan α = 2; cot β = 3
⇒ sec α = 5;cos ecβ = 10
∴ sec 2 α + cos ec 2β = 5 + 10 = 15
If tan −1 3 + tan −1 x = tan − 8 , then x =
1) 5
2) 1/5
Ans: 2
Sol. tan −1 3 + tan −1 x = tan −1 8
3+ x
1
=8⇒ x =
1 − 3x
5
9.
[EAMCET 2000]
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