Let α = tan ( 5π/16 sin ( 2cos ⁻¹( 1/√5 ) ) ) and β = cos ( sin ⁻¹( 4/5 ) + sec ⁻¹( 5/3 ) ) where the inverse trigonometric functions take principal values. Then, the equation whose roots are α and β is :

Solution

Maths sir · 4 steps

Turn every inverse function into one angle in a right triangle, and both roots fall out in two lines.

1. Find alpha. Let θ = cos⁻¹ 1/√5, so cos θ = 1/√5 and sin θ = 2/√5. Then sin 2θ = 2 × 2/√5 × 1/√5 = 4/5, so α = tan(5π/16 × 4/5) = tan π/4 = 1.
2. Same angle twice. Write sec⁻¹ 5/3 = cos⁻¹ 3/5. In the 3-4-5 triangle this is the same first-quadrant angle as A = sin⁻¹ 4/5. So β = cos 2A.

   β = cos 2A = 1 - 2 sin² A = 1 - 2 × 16/25 = -7/25

3. Build the quadratic. Sum of roots = 1 - 7/25 = 18/25 and product = -7/25. The equation is x² - 18/25x - 7/25 = 0, that is, 25x² - 18x - 7 = 0.
4. So. The roots are 1 and -7/25, and the equation is 25x² - 18x - 7 = 0, option (c).

The idea to keep. Convert each inverse trig value into an angle of a right triangle and read off sin, cos and tan from the sides. Before you expand anything, check whether two angles are secretly the same.

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