If tanØ + sinØ = m, tanØ - sinØ = n, find the value of m2 - n2
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
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.
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
If Cos x – Sin x = √2 Sin x, find the value of Cos x + Sin x:
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
If (2sinx) / (1 + cosx + Sinx) = t, (1 - Cosx + Sinx) / (1 + Sinx) can be written as:
= (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
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
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
Doc | 1 Page
Video | 18:42 min
Video | 09:43 min
Test | 10 questions | 20 min
Test | 5 questions | 4 min
Test | 25 questions | 25 min
Test | 15 questions | 15 min
Test | 10 questions | 10 min