Question -
Answer -
(i) cot x = 12/5, x in quadrant III
In third quadrant, tan x and cot x are positive. sinx, cos x, sec x, cosec x are negative.
By using the formulas,
tan x = 1/cot x
= 1/(12/5)
= 5/12
cosec x = –√(1 + cot2 x)
= –√(1 + (12/5)2)
= –√(25+144)/25
= –√(169/25)
= -13/5
sin x = 1/cosec x
= 1/(-13/5)
= -5/13
cos x = – √(1 – sin2 x)
= – √(1 – (-5/13)2)
= – √(169-25)/169
= – √(144/169)
= -12/13
sec x = 1/cos x
= 1/(-12/13)
= -13/12
∴ sin x = -5/13, cos x = -12/13,tan x = 5/12, cosec x = -13/5, sec x = -13/12
(ii) cos x = -1/2, x in quadrant II
In second quadrant, sin x and cosec x are positive.tan x, cot x, cos x, sec x are negative.
By using the formulas,
sin x = √(1 – cos2 x)
= √(1 – (-1/2)2)
= √(4-1)/4
= √(3/4)
= √3/2
tan x = sin x/cos x
= (√3/2)/(-1/2)
= -√3
cot x = 1/tan x
= 1/-√3
= -1/√3
cosec x = 1/sin x
= 1/(√3/2)
= 2/√3
sec x = 1/cos x
= 1/(-1/2)
= -2
∴ sin x = √3/2, tan x =-√3, cosec x = 2/√3, cot x = -1/√3 sec x = -2
(iii) tan x = 3/4, x in quadrant III
In third quadrant, tan x and cot x are positive. sinx, cos x, sec x, cosec x are negative.
By using the formulas,
sin x = √(1 – cos2 x)
= – √(1-(-4/5)2)
= – √(25-16)/25
= – √(9/25)
= – 3/5
cos x = 1/sec x
= 1/(-5/4)
= -4/5
cot x = 1/tan x
= 1/(3/4)
= 4/3
cosec x = 1/sin x
= 1/(-3/5)
= -5/3
sec x = -√(1 + tan2 x)
= – √(1+(3/4)2)
= – √(16+9)/16
= – √ (25/16)
= -5/4
∴ sin x = -3/5, cos x =-4/5, cosec x = -5/3, sec x = -5/4, cot x = 4/3
(iv) sin x = 3/5, x in quadrant I
In first quadrant, all trigonometric ratios arepositive.
So, by using the formulas,
tan x = sin x/cos x
= (3/5)/(4/5)
= 3/4
cosec x = 1/sin x
= 1/(3/5)
= 5/3
cos x = √(1-sin2 x)
= √(1 – (-3/5)2)
= √(25-9)/25
= √(16/25)
= 4/5
sec x = 1/cos x
= 1/(4/5)
= 5/4
cot x = 1/tan x
= 1/(3/4)
= 4/3
∴ cos x = 4/5, tan x =3/4, cosec x = 5/3, sec x = 5/4, cot x = 4/3