Test: Trigonometry- 1


5 Questions MCQ Test SSC CGL Tier 2 - Study Material, Online Tests, Previous Year | Test: Trigonometry- 1


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QUESTION: 1

If tanØ + sinØ = m, tanØ - sinØ = n, find the value of m2 - n2

Solution:

Adding the two equations,

► tanØ = (m + n) / 2

Subtracting the same,

► sinØ = (m - n) / 2

Since, there are no available direct formula for relation between sinØ tanØ but we know that

► cosec2Ø - cot2Ø = 1

∴(2 / (m - n)2 - (2 / m + n)2 = 1

► (4(m + n)2 - (m - n)2) / (m2 - n2)2 = 1
► (m2 - n2)2 = 4(4mn)
► m2 - n2 = 4√mn

QUESTION: 2

A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student.

Solution:

Let BC be the height of the tower and DC be the height of the student.
In rt. ΔABC

► AB = BC cot 45°
► AB = 100 m .......(1)

In rt. ΔABD

► AB = BD cot 60°
► AB = (BC + CD) cot 60°
► AB = (10 + CD) * 1 / √3

Equating (i) and (ii)

► (10 + CD) * 1 / √3 = 100
► (10 + CD) = 100√3
► CD = 100√3 - 100
= 10(1.732 - 1)
= 100 x 0.732
= 73.2 m

QUESTION: 3

If Cos x – Sin x = √2 Sin x, find the value of Cos x + Sin x:

Solution:

Cos x – Sin x = √2 Sin x
► Cos x = Sin x + √2 Sin x
► Cos x = Sin x + √2 Sin x
► Sin x = Cos x / (√2 + 1) * Cos x
► Sin x = (√2 - 1) / √2 - 1) * 1 / √2 + 1 * Cos x
► Sin x = (√2 - 1) / ((√2)2 - (1)2) * Cos x
► Sin x = (√2 - 1) Cos x
► Sin x = √2 Cos x - Cos x
► Sin x + Cos x = √2 Cos x

QUESTION: 4

If (2sinx) / (1 + cosx + Sinx) = t, (1 - Cosx + Sinx) / (1 + Sinx) can be written as:

Solution:

= (2Sinx) / (1 + cosx + Sinx) * (1 + Sinx - Cosx) / (1 + Sinx - Cosx) 

= 2 Sin x * (1 + Sinx - Cosx) / (1 + Sinx)2 - (Cos2x) 

= 2 Sin x * (1 + Sinx - Cosx) / (1 + Sin2x + 2Sinx - Cos2x)

= 2 Sin x . (1 + Sinx - Cosx) / (cos2x + Sin2x + Sin2x + 2Sinx - Cos2x)
(Since, 1 = Cos2x + Sin2x)

= 2 Sinx * (1 + Sinx - Cosx) / (2Sin2x + 2Sinx)

= 2 Sinx * (1 + Sinx - Cosx) / (2Sinx(1 + Sinx))

= (1 - Cosx + Sinx) / 1 + Sinx) / (1 + Sinx)
= t

QUESTION: 5

If cos A + cos2 A = 1 and a sin12 A + b sin10 A + c sin8 A + d sin6 A - 1 = 0. Find the value of a+b / c+d

Solution:

Correct Answer :- B

Explanation : Cos A = 1 - Cos2A

=> Cos A = Sin2A

=> Cos2A = Sin4A

=> 1 – Sin2A = Sin4A

=> 1 = Sin44A + Sin2A

=> 13 = (Sin4A + Sin2A)3

=> 1 = Sin12 A + Sin6A + 3Sin8A + 3Sin10A

=> Sin12A + Sin6A + 3Sin8A + 3Sin10A – 1 = 0

On comparing,

a = 1, b = 3 , c = 3 , d = 1

=  a+b/c+d

Hence, the answer is 1

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