Let S={θ ∈[0,2 π]: 8^(2 sin ² θ)+8^(2 cos ² θ)=16} . Then n(s) + Σ_(θ ∈ S) ( sec ( π/4 + 2θ )cos ec( π/4 + 2θ ) ) is equal to:

Solution

Maths sir · 4 steps

The two powers of 8 always multiply to 64, so put t = 8^(2 sin² θ) and get a perfect square.

1. Substitute. Let t = 8^(2 sin² θ). Since cos² θ = 1 - sin² θ, the second term is 8^(2 - 2 sin² θ) = 64/t. So the equation becomes t + 64/t = 16.

   t² - 16t + 64 = 0 ⇒ (t-8)² = 0 ⇒ t = 8

2. Find the set S. t = 8 means 2 sin² θ = 1, so sin² θ = 1/2. In [0, 2π] this gives θ = π/4, 3π/4, 5π/4, 7π/4, so n(S) = 4.
3. Simplify each term. For any angle x, sec x cosec x = 1/(sin x cos x) = 2/(sin 2x). With x = π/4 + 2θ we get sin 2x = sin(π/2 + 4θ) = cos 4θ, so each term is 2/(cos 4θ).

   θ	4θ	cos 4θ	Term 2/(cos 4θ)
   π/4	π	-1	-2
   3π/4	3π	-1	-2
   5π/4	5π	-1	-2
   7π/4	7π	-1	-2

4. So. The sum of the four terms is -8, and n(S) + (-8) = 4 - 8 = -4, which is option (c).

The idea to keep. When two exponents with the same base add up to a constant, put t equal to one power and you get a quadratic. Also remember that sec x cosec x = 2/(sin 2x).

Answer from the NTA final answer key. Solution by PrepWiser.