1 |
cos-1(cos x) = |
x
cos x
x = 1/x
cos<sup>-2</sup> x
|
2 |
Cos-1(x)= |
cos x
x
tan-1(-x)
Sec-1 (1/x)
|
3 |
Cos-1 (-x) = |
-x
1/x
tan-1 x
π-cos-1 x
|
4 |
Ifπ≤x≤2π, then cos-1 (cos x)= |
cos x
-x
1/x
-x
|
5 |
If cos (2 sin-1 x) = 1/9 , then what is the value of x? |
1/3
-2/3
2/3
2/3 , -2/3
|
6 |
Cos (cos4π/3)= |
π/2
π/3
2π/3
-π/3
|
7 |
The exact degree value of the function sin-1( -√3/2) is |
70<sup>ο</sup>
50<sup>ο</sup>
90<sup>ο</sup>
60<sup>ο</sup>
|
8 |
What is the value of cos (cos-1 2) ? |
√2
1/2
undefine
0
|
9 |
The value of cos(cos-1 1/2) is |
1/2
√3/2
-1/2
1/√2
|
10 |
What is the value of cos-1(1/2)? |
π/3
π/4
3π/2
π/6
|
11 |
sin-1 x = |
tan<sup>-1</sup> x
Cosec<sup> -1</sup> x
Cosec x
cosec<sup>-1</sup>(1/x)
|
12 |
Sin-1(-x)= |
x
-x
-sin-1 x
cos-1 x
|
13 |
sin -1(sin2π/3) = |
π/2
2π/3
-3π/2
π/3
|
14 |
sn (2sin-10.8) |
0.56
0.69
-0.16
0.96
|
15 |
Sin -1 x= |
sin(π/2-x)
Sin-1 (π/2-x)
π/2-cos-1x
π/2 + cos-1x
|
16 |
sin (sin-1(1/2))= |
0
2
∞
1/2
|
17 |
The principal value of sin-1[-√(√3) /2] is |
5π/3
-2π/3
-<img width="9" height="19" src="file:///C:/Users/Softsol/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png" v:shapes="_x0000_i1025">π/3<p class="MsoNormal"><!--[endif]--><o:p></o:p></p>
π/3
|
18 |
The value of sin-1 24/25 is equal to |
csc-1 25/24
sec-1 24/25
2 tan-1 4/5
2cos-1 24/25
|
19 |
The value of sin-1 5/13 is equal to |
Cos 5/13
Tan<sup>-1</sup> 5/12
cos<sup>-1</sup>5/12
2 cos<sup>-1 </sup>4/5
|
20 |
The Principal value of sin-1 (-1/1/2) |
<p class="MsoNormal">π/2<o:p></o:p></p>
<p class="MsoNormal">-π/2<o:p></o:p></p>
<p class="MsoNormal">π<o:p></o:p></p>
<p class="MsoNormal">-π<o:p></o:p></p>
|
21 |
In the interval 0≤ x ≤ π, the sine is
|
Not a function
Not defined
Infinity
Not one-to-one function
|
22 |
x = sin-1 3, then the value of sin x is |
√(3/2)
3
Not possible
-1
|
23 |
The domain of the function y = sin x , is |
-π/2≤ x≤ π/2
π/ ≤ x≤ π
-2π ≤ x≤ 2π
-1 ≤ x≤ 1
|
24 |
The principal value of sin-1 (-1/2) |
π/3
π/4
π/6
-π/6
|
25 |
The principal value of sin-1√(3/2) is |
-π/3
π/3
2π/3
π/2
|
26 |
The law of sines can be used to solve oblique triangle when following information is given: |
Two angles and a side
Two sides and an angle opposite one of the given sides
Two sides and the angle between two sided
Option a and b
|
27 |
The law of sines can be used to solve |
Right angle triangle
Isosceles triangle
oblique triangle
haxagon
|
28 |
If sided of𝜟ABC are 16,20,and 33, then the value of the greatests angle to |
150𝜊 20'
132𝜊 35'
101𝜊 25'
160𝜊 50'
|
29 |
IfΔABC is right, law of cosine reduce to |
Law of sine
Law of tangent
Phthogorous theorem
Hero's formula
|
30 |
In triangle ABC, in which b=95, c=34, a =52𝜊then the value of a= |
18 cm
18.027 cm
20.7 cm
19 cm
|