Chapter 5.2 Trigonometric Functions Answers

In Practise 5.two of Chapter v, we shall discuss problems based on signs of trigonometric functions and variations in values of trigonometric functions in different quadrants. Students can refer to RD Sharma Class eleven Solutionswhich is prepared by our skillful tutors to assist students clarify their doubts pertaining to this chapter. On regular practice, students can speed up the method of solving problems by using brusque-cut tips to secure high marks in their test. Students tin can download the pdf hands from the links given beneath.

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Exercise 5.1

Practise v.3

Access answers to RD Sharma Solutions for Course 11 Maths Practice 5.two Affiliate 5 – Trigonometric Functions

1. Find the values of the other v trigonometric functions in each of the following:

(i) cot x = 12/5, x in quadrant Iii

(ii) cos x = -1/two, x in quadrant Two

(iii) tan 10 = 3/4, x in quadrant Iii

(four) sin x = 3/v, 10 in quadrant I

Solution:

(i) cot 10 = 12/5, ten in quadrant III

In third quadrant, tan x and cot x are positive. sin x, cos x, sec 10, cosec 10 are negative.

By using the formulas,

tan ten = 1/cot x

= ane/(12/5)

= 5/12

cosec x = –√(one + cot² 10)

= –√(1 + (12/5)²)

= –√(25+144)/25

= –√(169/25)

= -13/five

sin x = ane/cosec x

= 1/(-13/v)

= -5/thirteen

cos x = – √(1 – sin² x)

= – √(1 – (-v/13)²)

= – √(169-25)/169

= – √(144/169)

= -12/13

sec x = 1/cos ten

= ane/(-12/13)

= -13/12

∴ sin x = -5/13, cos x = -12/13, tan 10 = 5/12, cosec ten = -13/five, sec x = -13/12

(ii) cos x = -i/two, x in quadrant II

In second quadrant, sin x and cosec x are positive. tan x, cot x, cos x, sec x are negative.

Past using the formulas,

sin 10 = √(1 – cos² x)

= √(1 – (-i/2)ⁱⁱ)

= √(4-i)/4

= √(three/iv)

= √3/2

tan x = sin x/cos x

= (√3/2)/(-1/two)

= -√iii

cot x = 1/tan x

= 1/-√3

= -ane/√3

cosec x = i/sin x

= ane/(√3/2)

= two/√3

sec x = 1/cos ten

= 1/(-1/2)

= -two

∴ sin x = √3/2, tan ten = -√3, cosec x = 2/√3, cot x = -1/√3 sec x = -two

(three) tan x = iii/four, x in quadrant III

In 3rd quadrant, tan x and cot x are positive. sin x, cos x, sec 10, cosec 10 are negative.

By using the formulas,

sin ten = √(1 – cos² x)

= – √(1-(-4/five)²)

= – √(25-xvi)/25

= – √(9/25)

= – iii/five

cos x = 1/sec x

= i/(-5/4)

= -4/five

cot x = 1/tan x

= 1/(three/4)

= four/iii

cosec ten = one/sin x

= one/(-iii/5)

= -5/3

sec x = -√(1 + tanⁱⁱ 10)

= – √(1+(3/4)²)

= – √(16+9)/16

= – √ (25/16)

= -v/4

∴ sin x = -3/5, cos x = -4/5, cosec x = -5/three, sec x = -v/4, cot 10 = four/3

(iv) sin x = 3/5, x in quadrant I

In first quadrant, all trigonometric ratios are positive.

So, by using the formulas,

tan x = sin 10/cos x

= (3/five)/(four/5)

= 3/four

cosec x = 1/sin x

= 1/(three/5)

= five/3

cos 10 = √(1-sin² x)

= √(i – (-iii/5)ⁱⁱ)

= √(25-9)/25

= √(16/25)

= 4/5

sec 10 = 1/cos x

= i/(4/v)

= five/4

cot 10 = 1/tan x

= i/(3/4)

= 4/iii

∴ cos x = 4/5, tan x = 3/four, cosec 10 = 5/3, sec x = v/4, cot x = 4/3

2. If sin x = 12/thirteen and lies in the second quadrant, discover the value of sec x + tan x.

Solution:

Given:

Sin 10 = 12/thirteen and x lies in the 2d quadrant.

We know, in second quadrant, sin x and cosec ten are positive and all other ratios are negative.

Past using the formulas,

Cos x = √(1-sin² x)

= – √(1-(12/thirteen)ⁱⁱ)

= – √(one- (144/169))

= – √(169-144)/169

= -√(25/169)

= – five/13

We know,

tan x = sin ten/cos x

sec x = one/cos x

At present,

tan x = (12/thirteen)/(-5/13)

= -12/5

sec x = 1/(-5/13)

= -13/5

Sec x + tan x = -13/v + (-12/5)

= (-13-12)/5

= -25/five

= -5

∴ Sec 10 + tan 10 = -5

iii. If sin x = iii/5, tan y = 1/ii and π/2 < x< π< y< 3π/2 find the value of 8 tan x -√5 sec y.

Solution:

Given:

sin x = three/5, tan y = 1/2 and π/two < x< π< y< 3π/2

We know that, x is in second quadrant and y is in third quadrant.

In second quadrant, cos x and tan x are negative.

In third quadrant, sec y is negative.

Past using the formula,

cos 10 = – √(1-sin² x)

tan x = sin x/cos 10

Now,

cos ten = – √(1-sin² ten)

= – √(1 – (3/5)ⁱⁱ)

= – √(one – 9/25)

= – √((25-ix)/25)

= – √(16/25)

= – 4/5

tan ten = sin x/cos x

= (three/5)/(-4/5)

= iii/5 × -5/4

= -3/4

We know that sec y = – √(one+tan² y)

= – √(1 + (1/two)²)

= – √(1 + ane/iv)

= – √((4+one)/iv)

= – √(5/4)

= – √v/2

At present, 8 tan 10 – √5 sec y = 8(-3/4) – √v(-√five/2)

= -6 + v/2

= (-12+five)/2

= -7/ii

∴ 8 tan x – √5 sec y = -vii/2

4. If sin x + cos ten = 0 and 10 lies in the 4th quadrant, observe sin ten and cos x.

Solution:

Given:

Sin x + cos x = 0 and ten lies in 4th quadrant.

Sin x = -cos x

Sin x/cos x = -1

So, tan ten = -1 (since, tan x = sin x/cos ten)

Nosotros know that, in 4th quadrant, cos ten and sec x are positive and all other ratios are negative.

By using the formulas,

Sec 10 = √(i + tan² x)

Cos x = 1/sec x

Sin x = – √(one- cos^two x)

Now,

Sec x = √(ane + tan² x)

= √(one + (-1)²)

= √2

Cos 10 = 1/sec ten

= 1/√2

Sin 10 = – √(1 – cosⁱⁱ x)

= – √(ane – (1/√2)^two)

= – √(one – (1/2))

= – √((ii-1)/ii)

= – √(ane/ii)

= -1/√2

∴ sin 10 = -one/√two and cos x = i/√2

5. If cos x = -3/5 and π<x<3π/2 discover the values of other five trigonometric functions and hence evaluate

Solution:

Given:

cos ten= -three/five and π <x < 3π/2

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

By using the formulas,

Sin ten = – √(1-cos² x)

Tan ten = sin x/cos x

Cot x = i/tan ten

Sec x = ane/cos x

Cosec ten = i/sin x

Now,

Sin 10 = – √(1-cos² x)

= – √(1-(-3/5)^two)

= – √(1-ix/25)

= – √((25-9)/25)

= – √(xvi/25)

= – four/v

Tan x = sin x/cos x

= (-iv/five)/(-three/5)

= -4/five × -v/iii

= four/3

Cot 10 = 1/tan x

= 1/(4/iii)

= 3/iv

Sec x = 1/cos x

= one/(-3/5)

= -v/3

Cosec ten = one/sin x

= 1/(-four/5)

= -five/4

∴
= [(-5/four) + (3/4)] / [(-5/3) – (4/3)]

= [(-five+three)/iv] / [(-5-4)/three]

= [-2/4] / [-nine/three]

= [-i/ii] / [-3]

= ane/6