RD Chapter 5 Trigonometric Functions Ex 5.2 Solutions
Question - 1 : - Find the values of the other five trigonometric functions in each of the following:
(i) cot x = 12/5, x in quadrant III
(ii) cos x = -1/2, x in quadrant II
(iii) tan x = 3/4, x in quadrant III
(iv) sin x = 3/5, x in quadrant I
Answer - 1 : -
(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
Question - 2 : - If sin x = 12/13 and lies in the second quadrant, find the value of sec x + tan x.
Answer - 2 : -
Given:
Sin x = 12/13 and x lies in the second quadrant.
We know, in second quadrant, sin x and cosec x arepositive and all other ratios are negative.
By using the formulas,
Cos x = √(1-sin2 x)
= – √(1-(12/13)2)
= – √(1- (144/169))
= – √(169-144)/169
= -√(25/169)
= – 5/13
We know,
tan x = sin x/cos x
sec x = 1/cos x
Now,
tan x = (12/13)/(-5/13)
= -12/5
sec x = 1/(-5/13)
= -13/5
Sec x + tan x = -13/5 + (-12/5)
= (-13-12)/5
= -25/5
= -5
∴ Sec x + tan x = -5
Question - 3 : - If sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 find the value of 8 tan x -√5 sec y.
Answer - 3 : -
Given:
sin x = 3/5, tan y = 1/2 and π/2 < x< π
We know that, x is in second quadrant and y is inthird quadrant.
In second quadrant, cos x and tan x are negative.
In third quadrant, sec y is negative.
By using the formula,
cos x = – √(1-sin2 x)
tan x = sin x/cos x
Now,
cos x = – √(1-sin2 x)
= – √(1 – (3/5)2)
= – √(1 – 9/25)
= – √((25-9)/25)
= – √(16/25)
= – 4/5
tan x = sin x/cos x
= (3/5)/(-4/5)
= 3/5 × -5/4
= -3/4
We know that sec y = – √(1+tan2 y)
= – √(1 + (1/2)2)
= – √(1 + 1/4)
= – √((4+1)/4)
= – √(5/4)
= – √5/2
Now, 8 tan x – √5 sec y = 8(-3/4) – √5(-√5/2)
= -6 + 5/2
= (-12+5)/2
= -7/2
∴ 8 tan x – √5 sec y =-7/2
Question - 4 : - If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x.
Answer - 4 : -
Given:
Sin x + cos x = 0 and x lies in fourth quadrant.
Sin x = -cos x
Sin x/cos x = -1
So, tan x = -1 (since, tan x = sin x/cos x)
We know that, in fourth quadrant, cos x and sec x arepositive and all other ratios are negative.
By using the formulas,
Sec x = √(1 + tan2 x)
Cos x = 1/sec x
Sin x = – √(1- cos2 x)
Now,
Sec x = √(1 + tan2 x)
= √(1 + (-1)2)
= √2
Cos x = 1/sec x
= 1/√2
Sin x = – √(1 – cos2 x)
= – √(1 – (1/√2)2)
= – √(1 – (1/2))
= – √((2-1)/2)
= – √(1/2)
= -1/√2
∴ sin x = -1/√2 and cos x= 1/√2
Question - 5 : - If cos x = -3/5 and π