Given that cot x = 5/ sqrt 11 find csc x
A - 5/6
B - 6/5
C - sqrt11/ 6
D - 6/ sqrt11
if cot x = 5/√11 , then
tan x = √11/5
and x must be in quads I or III
construct your right-angled triangle with opposite to angle x as √11 and adjacent as 5
r^2 = 11 + 25 = 36
r = 6
then cscx = 6/√11 ins quad I
csc x = -6√11 in III
your choices are missing the quad III angle.
draw the triangle. Review your basic trig functions:
opposite = √11
adjacent = 5
hypotenuse = 6
csc = hypotenuse/opposite
ctg x = cos x / sin x
cot²x = cos² x / sin² x
cot²x = ( 1 - sin² x ) / sin² x
cot²x = 1 / sin² x - sin x / sin² x
cot²x = csc² x - 1
Add 1 to both sides
cot²x + 1 = csc² x - 1 + 1
cot²x + 1 = csc² x
csc² x = cot²x + 1
csc x = ±√( 1 + cot² x )
csc x = ±√[ 1 + ( 5 / √11 )² ]
csc x = ±√[ 1 + 5² / ( √11 )² ]
csc x = ±√( 1 + 25 / 11 )
csc x = ±√( 11 / 11 + 25 / 11 )
csc x = ±√( 36 / 11 )
csc x = ±√36 / √11
csc x = ± 6 / √11
In your list only csc x = - 6 / √11 is correct.
In this case x lie in quadrant III where cotangent is positive and cosecant is negative.
My typo.
cot²x = 1 / sin² x - sin² x / sin² x
To find csc x (cosecant of x), we need to use the trigonometric identity: csc x = 1/ sin x.
Given that cot x = 5/√11, let's use the Pythagorean Identity to find sin x:
cot^2 x + 1 = csc^2 x
(5/√11)^2 + 1 = csc^2 x
(25/11) + 1 = csc^2 x
36/11 = csc^2 x
Taking the square root of both sides gives:
csc x = √(36/11)
To simplify this further, we can write it as:
csc x = (√36) / (√11)
csc x = 6 / √11
So, the answer is option D - 6/√11.