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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 = тАУтИЪ(1 + cot2 x)

= тАУтИЪ(1 + (12/5)2)

= тАУтИЪ(25+144)/25

= тАУтИЪ(169/25)

= -13/5

sin x = 1/cosec x

= 1/(-13/5)

= -5/13

cos x = тАУ тИЪ(1 тАУ sin2 x)

= тАУ тИЪ(1 тАУ (-5/13)2)

= тАУ тИЪ(169-25)/169

= тАУ тИЪ(144/169)

= -12/13

sec x = 1/cos x

= 1/(-12/13)

= -13/12

тИ┤ 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 = тИЪ(1 тАУ cos2 x)

= тИЪ(1 тАУ (-1/2)2)

= тИЪ(4-1)/4

= тИЪ(3/4)

= тИЪ3/2

tan x = sin x/cos x

= (тИЪ3/2)/(-1/2)

= -тИЪ3

cot x = 1/tan x

= 1/-тИЪ3

= -1/тИЪ3

cosec x = 1/sin x

= 1/(тИЪ3/2)

= 2/тИЪ3

sec x = 1/cos x

= 1/(-1/2)

= -2

тИ┤ sin x = тИЪ3/2, tan x =-тИЪ3, cosec x = 2/тИЪ3, cot x = -1/тИЪ3 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 = тИЪ(1 тАУ cos2 x)

= тАУ тИЪ(1-(-4/5)2)

= тАУ тИЪ(25-16)/25

= тАУ тИЪ(9/25)

= тАУ 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 = -тИЪ(1 + tan2 x)

= тАУ тИЪ(1+(3/4)2)

= тАУ тИЪ(16+9)/16

= тАУ тИЪ (25/16)

= -5/4

тИ┤ 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 = тИЪ(1-sin2 x)

= тИЪ(1 тАУ (-3/5)2)

= тИЪ(25-9)/25

= тИЪ(16/25)

= 4/5

sec x = 1/cos x

= 1/(4/5)

= 5/4

cot x = 1/tan x

= 1/(3/4)

= 4/3

тИ┤ 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 = тИЪ(1-sin2 x)

= тАУ тИЪ(1-(12/13)2)

= тАУ тИЪ(1- (144/169))

= тАУ тИЪ(169-144)/169

= -тИЪ(25/169)

= тАУ 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

тИ┤ Sec x + tan x = -5

Question - 3 : - If sin x = 3/5, tan y = 1/2 and ╧А/2 < x< ╧А< y< 3╧А/2 find the value of 8 tan x -тИЪ5 sec y.

Answer - 3 : -

Given:

sin x = 3/5, tan y = 1/2 and ╧А/2 < x< ╧А

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 = тАУ тИЪ(1-sin2 x)

tan x = sin x/cos x

Now,

cos x = тАУ тИЪ(1-sin2 x)

= тАУ тИЪ(1 тАУ (3/5)2)

= тАУ тИЪ(1 тАУ 9/25)

= тАУ тИЪ((25-9)/25)

= тАУ тИЪ(16/25)

= тАУ 4/5

tan x = sin x/cos x

= (3/5)/(-4/5)

= 3/5 ├Ч -5/4

= -3/4

We know that sec y = тАУ тИЪ(1+tan2 y)

= тАУ тИЪ(1 + (1/2)2)

= тАУ тИЪ(1 + 1/4)

= тАУ тИЪ((4+1)/4)

= тАУ тИЪ(5/4)

= тАУ тИЪ5/2

Now, 8 tan x тАУ тИЪ5 sec y = 8(-3/4) тАУ тИЪ5(-тИЪ5/2)

= -6 + 5/2

= (-12+5)/2

= -7/2

тИ┤ 8 tan x тАУ тИЪ5 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 = тИЪ(1 + tan2 x)

Cos x = 1/sec x

Sin x = тАУ тИЪ(1- cos2 x)

Now,

Sec x = тИЪ(1 + tan2 x)

= тИЪ(1 + (-1)2)

= тИЪ2

Cos x = 1/sec x

= 1/тИЪ2

Sin x = тАУ тИЪ(1 тАУ cos2 x)

= тАУ тИЪ(1 тАУ (1/тИЪ2)2)

= тАУ тИЪ(1 тАУ (1/2))

= тАУ тИЪ((2-1)/2)

= тАУ тИЪ(1/2)

= -1/тИЪ2

тИ┤ sin x = -1/тИЪ2 and cos x= 1/тИЪ2

Question - 5 : - If cos x = -3/5 and ╧А

Answer - 5 : -

Given:

cos x= -3/5 and ╧А

We know that in the third quadrant, tan x and cot xare positive and all other rations are negative.

By using the formulas,

Sin x = тАУ тИЪ(1-cos2 x)

Tan x = sin x/cos x

Cot x = 1/tan x

Sec x = 1/cos x

Cosec x = 1/sin x

Now,

Sin x = тАУ тИЪ(1-cos2 x)

= тАУ тИЪ(1-(-3/5)2)

= тАУ тИЪ(1-9/25)

= тАУ тИЪ((25-9)/25)

= тАУ тИЪ(16/25)

= тАУ 4/5

Tan x = sin x/cos x

= (-4/5)/(-3/5)

= -4/5 ├Ч -5/3

= 4/3

Cot x = 1/tan x

= 1/(4/3)

= 3/4

Sec x = 1/cos x

= 1/(-3/5)

= -5/3

Cosec x = 1/sin x

= 1/(-4/5)

= -5/4

тИ┤ 
= [(-5/4) + (3/4)] / [(-5/3) тАУ (4/3)]

= [(-5+3)/4] / [(-5-4)/3]

= [-2/4] / [-9/3]

= [-1/2] / [-3]

= 1/6

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