Let S={x ∈(-π/2, π/2): 9^(1-tan ² x)+9^(tan ² x)=10} and β=Σx ∈ S tan ²(x/3), then 1/6(β-14)² is equal to :

Solution

Maths sir · 4 steps

Put t = 9^(tan² x) and the equation becomes a quadratic in t.

1. Substitute. Write 9^(1-tan² x) = 9/9^(tan² x) and let t = 9^(tan² x). The equation becomes 9/t + t = 10, that is t² - 10t + 9 = 0, so (t-1)(t-9) = 0.
2. Back to x. t = 1 gives tan² x = 0, and t = 9 gives tan² x = 1. In (-π/2, π/2) this means x = 0 or x = ±π/4, so S = {0, π/4, -π/4}.
3. Find beta. The values of x/3 are 0 and ±π/12. Since tan π/12 = 2 - √3, we get tan² π/12 = 7 - 4√3, and it counts twice.

   β = 0 + 2(7 - 4√3) = 14 - 8√3

4. So. β - 14 = -8√3, so (β - 14)² = 192 and 1/6 × 192 = 32. The answer is (b).

The idea to keep. When the exponents add up to a constant, like 1 - tan² x and tan² x, set t equal to one power and the other becomes a constant over t. Then count every root in the interval, including the negative ones.

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