SIN & COS (SINE & COSINE RULES)

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Trigonometry (SIN and COSINE Rules) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

SHIV KISHOR

QUERY 1

If cosx + secx = 3; then (tanx)² – (sinx)² =?

A) 4
B) 5
C) 13
D) 8

Atul Roy
tanx² ­- sinx²

= (secx² ­- 1) ­- (1 ­- cosx²)

= secx² + cosx² ­- 2

= (secx + cosx)² ­- 2secx- cosx ­- 2

= 3² ­- 2 ­- 2

= 9 ­- 4

= 5 (option ‘B’)

 

QUERY 2

If tanx*tan²x = 1; then sin² 2x + tan² 2x = ?

a) 3/4
b) 10/3
c) 15/4
d) 3

sukhjinder dhiman
tanx*tan²x = 1
=> tanx = 1/tan²x
=> tanx = cot²x
=> tanx = tan (90 ­- 2x)
=> x = 90 ­- 2x
=> 3x = 90
=> x = 30

Put this value in (sin² x)² + (tan² x)²
= (sin 2*30)² + (tan 2*30)²
= (sin 60)² + (tan 60)²
= (√3/2)² + (√3)²
= 3/4 + 3
= 15/4 (option ‘C’)

QUERY 3

If 2 sin X + 15 cos² X = 7; then the value of tan X is?

A) 4/3 or -2√5/5
B) 3/4 0r -2√5/8
C) 2/√5 0r 3/4
D) √5 0r -2√5/8

Sumanashis
2sin X + 15cos² X + 7
=> 2sin X + 15 (1 – sin² X) = 7
=> 2sin X = 7 – 15 + 15 sin² X
=> 15sin² X – 2sin X – 8 = 0
=> 15sin² X – 12sin X + 10sin X – 8 = 0
=> 3sin X (5sin X – 4) + 2 (5sin X – 4) = 0
=> sin X = -2/3 or sin X = 4/5

For sin X = -2/3; hypotenuse = 3 and the sides are -2 and √5. For sin X = 4/5, we can say that one side of the triangle is 4 and the hypotenuse is 5; so the other side must be 3 as 3, 4 & 5 is a triplet.

Hence cos X = 3/5 or √5/3
Now tan X = sin X/cos X = (4/5)*(5/3) = 4/3 or (-2/3)*(3/√5) = -2/√5 = -2√5/5
So tan X = 4/3 or -2√5/5 (option ‘A’)

QUERY 4

If cosec X – sin X = a³ and sec X – cos X = b³; then what is the value of a²b²(a² + b²)

A) 0
B) 1
C) 2
D) 3

Vijay Bharath Reddy
a³ = cosec X – sin X and b³ = sec X – cos X (given)

=> a³ = 1/sin X – sin X and b³ = 1/cos X – cos X

=> a³ = (1 – sin² X)/sin X and b³ = (1 – cos² X)/cos X

=> a³ = (cos² X/sin X) and b³ = (sin² X/cos X)

=> a = ∛(cos² X/sin X) and b = ∛(sin² X/cos X)

Now the given expression
a²b²(a² + b²)
= a⁴b² + a²b⁴

Now by putting the values of a and b in it, the expression
= [(cos² X/sin X)^4/3][(sin² X/cos X)^2/3] + [(cos² X/sin X)^2/3][(sin² X/cos X)^4/3]

= cos² X + sin² X
= 1 (option ‘B’)

QUERY 5

If (a² – b²) sin X + 2ab cos X = a² + b²; then tan X = ?

A) (a² + b²)/2ab
B) (a² – b²)/2
C) (a² – b²)/2ab
D) (a² + b²)/2

pawan ,,,
(a² – b²) sin X + 2ab cos X = a² + b²

=> a² sin X – b² sin X – 2ab cos X = a² + b²

=> a² sin X – b² sin X – 2ab cos X – a² – b² = 0

=> -a² (1 – sin X) – b² (1 + sin X) – 2ab cos X = 0

=> a² (1 – sin X) + b² (1 + sin X) + 2ab cos X = 0 (By change of signs)

=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab cos X = 0

=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√cos² X

=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√(1 – sin² X)

=> [a√(1 – sin X)]² + [b√(1 + sin X)]² + 2ab√[(1 + sin X) (1 – sin X)]

=> [(a√(1 – sin X) + b√(1 + sin X)]² = 0

=> a√(1 – sin X) = -b√(1 + sin X)

=> a²(1 – sin X) = b²(1 + sin X)

=> (a² + b²) sin X = a² – b²

=> sin X = (a² – b²)/(a² + b²)

=> cos X = 2ab/(a² + b²) [cos X = √(1 -sin² X)]

=> tan X = (a² – b²)/(2ab) (option ‘C’)

QUERY 6

If sin A + cos A = 17/13, 0 < x < 90; then sin A – cos A is ?

A) 5/17
B) 3/19
C) 7/10
D) 7/13

RONNIE BANSAL
Squaring sin A + cos A = 17/13 on both sides
sin² A + cos² A + 2sin A cos A = 289/169

=> 2sin A cos A = 289/169 – 1                              (by putting sin² A + cos² A = 1)

=> 2sin A cos A = 120/169

=> sin A cos A = 60/169

Now sin A – cos A
=√(sin A – cos A)²

= √(sin² A + cos² A – 2 sin A cos A)²

= √[1 – 2*(60/169)]²                                               (by putting sin² A + cos² A = 1 & sin A cos A = 80/169)

= √(49/169)

= 7/13 (option ‘D’)

QUERY 7

If tan² A = 2tan² B + 1; then find cos² A + sin² B?

A) 0
B) 1
C) 2
D) -1

JAYANTCHARAN CHARAN
Since both the terms in the “find” entry are in squares ………………….so the answer cannot be -1,

again it cannot be 0 also……..because than both the terms should be zero as none can be negative for both the terms to be 0.

again the answer cannot be 2 also because then both the terms in FIND entry should be 1 as it will make tan 90 which is undefined

so the correct answer is 1 one only (option ‘B’)

QUERY 8

If cot A + cosec A = 3 and A is an acute angle; then the value of cos A is ?

A) -1
B) -3/5
C) -2/3
D) 4/5

MAHA GUPTA
cosec A + cot A = 3 (given)                                                                     —- (i)

cosec² A – cot² A = 1
=> (cosec A + cot A)(cosec A – cot A) = 1
=> cosec A – cot A) =¹⁄₃             (by putting cosec A + cot A = 3)        —- (ii)

By adding (i) and (ii)
2cosec A = ¹⁰⁄₃
=> cosec A = ¹⁰⁄₆

By Subtracting (ii) from (i)
2 cot A = ⁸⁄₃
=> cot A = ⁸⁄₆

cos A = ^{cot A}⁄_{cosec A} = ^(8/6)⁄_(10/6)
= ⁴⁄₅ (option ‘D’)

QUERY 9

SHIV KISHOR
(tan 27 + cot 63)/[tan 27(sin 25 + cos 65)]
= (tan 27 + tan 27)/[tan 27(sin 25 + sin 25)]
= 2tan 27/(tan27*2sin25)
= 2/2sin25
= 1/sin25
= cosec 25 (option ‘D’)

QUERY 10

(sin 10)(sin 30)(sin 50)(sin 70) = ?

A) 16
B) 7
C) 1/16
D) 1/4

SHIV KISHOR
sin10*sin30*sin50* sin70
= cos80*cos60*cos40*cos20
= cos60*[cos20*cos(2*20)*cos(4*20)]
= cos60*[(1/4)cos60]
= (1/2)*(1/4)*(1/2)
= 1/16 (option ‘D’)