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QUESTION: 1

If tanØ + sinØ = m, tanØ - sinØ = n, find the value of **m ^{2} - n^{2}**

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

► cosec^{2}Ø - cot^{2}Ø = 1

∴(2 / (m - n)^{2} - (2 / m + n)^{2} = 1

► (4(m + n)^{2} - (m - n)^{2}) / (m^{2} - n^{2})^{2} = 1

► (m^{2} - n^{2})^{2} = 4(4mn)

► m^{2} - n^{2} = 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} - (Cos^{2}x)

= 2 Sin x * (1 + Sinx - Cosx) / (1 + Sin^{2}x + 2Sinx - Cos^{2}x)

= 2 Sin x . (1 + Sinx - Cosx) / (cos^{2}x + Sin^{2}x + Sin^{2}x + 2Sinx - Cos^{2}x)

(Since, 1 = Cos^{2}x + Sin^{2}x)

= 2 Sinx * (1 + Sinx - Cosx) / (2Sin^{2}x + 2Sinx)

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

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

= t

QUESTION: 5

If cos A + cos^{2} A = 1 and a sin^{12} A + b sin^{10} A + c sin^{8} A + d sin^{6} A - 1 = 0. Find the value of a+b / c+d

Solution:

**Correct Answer :- B**

**Explanation : **Cos A = 1 - Cos^{2}A

=> Cos A = Sin^{2}A

=> Cos^{2}A = Sin^{4}A

=> 1 – Sin^{2}A = Sin^{4}A

=> 1 = Sin4^{4}A + Sin^{2}A

=> 1^{3} = (Sin^{4}A + Sin^{2}A)^{3}

=> 1 = Sin^{12} A + Sin^{6}A + 3Sin^{8}A + 3Sin^{10}A

=> Sin^{12}A + Sin^{6}A + 3Sin^{8}A + 3Sin^{10}A – 1 = 0

On comparing,

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

= a+b/c+d

Hence, the answer is 1

### Maths test :-Some applications of trigonometry.

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