Trigonometric EquationHard
Question
For a positive integer n let (fn(θ)
(1 + sec θ)(1 + sec 2θ)(1 + sec 22θ)...(1+ sec 2nθ), then
(1 + sec θ)(1 + sec 2θ)(1 + sec 22θ)...(1+ sec 2nθ), thenOptions
A.f2
= 1
= 1B.f3
= 1
= 1C.f4
= 1
= 1D.f5
= 1
= 1Solution
Note Multiplicative loop is very important approach in IIT Mathematics.
Now,
(1 + sec θ) = 
0
∴ fn(θ) = (tan θ / 2)(1 + sec θ)
(1 + sec 2θ)(1 + sec 22θ)...(1+ sec 2nθ)
= (tan θ)(1 + sec 2θ)(1 + sec 22θ)...(1 + sec 2nθ)
= tan 2θ.(1 + sec 22θ)...(1 + sec 2nθ)
.........................................................
.........................................................
= tan(2nθ)
Now,
Therefore, (a) is the answer.
Therefore, (b) is the answer.
Therefore, (c) is the answer.
Therefore, (d) is the answer.
Now,
(1 + sec θ) = 0 ∴ fn(θ) = (tan θ / 2)(1 + sec θ)
(1 + sec 2θ)(1 + sec 22θ)...(1+ sec 2nθ)
= (tan θ)(1 + sec 2θ)(1 + sec 22θ)...(1 + sec 2nθ)
= tan 2θ.(1 + sec 22θ)...(1 + sec 2nθ)
.........................................................
.........................................................
= tan(2nθ)
Now,
Therefore, (a) is the answer.
Therefore, (b) is the answer.
Therefore, (c) is the answer.
Therefore, (d) is the answer.
Create a free account to view solution
View Solution FreeMore Trigonometric Equation Questions
The equation of a normal to the curve, sin y = x sin at x = 0, is :...If α + β = and β + γ = α, then tan α equals...Let two non-collinear unit vectors and form an acute angle. A point P moves so that at any time t the positionvector (wh...Let I = , J = . Then, for an arbitrary constant C, the value of J - I equals......