Test: Trigonometry- 1

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

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

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

= (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

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