50 tan ( 3tan ⁻¹( 1/2 ) + 2cos ⁻¹( 1/√5 ) ) + 4√2 tan ( 1/2tan ⁻¹(2√2 ) ) is equal to ____________.

Solution

Maths sir · 4 steps

Spot the complement first: cos⁻¹ 1/√5 = π/2 - tan⁻¹ 1/2.

1. Name the angle. Let α = tan⁻¹ 1/2. Draw a right triangle with legs 1 and 2 and hypotenuse √5. The other acute angle, opposite the side 2, has cosine 1/√5, so cos⁻¹ 1/√5 = π/2 - α.

   3α + 2(π/2 - α) = π + α

2. First term. tan has period π, so tan(π + α) = tan α = 1/2. The first term is 50 × 1/2 = 25.
3. Second term. Let θ = tan⁻¹(2√2). This is an acute angle with cos θ = 1/3 and sin θ = 2√2/3. Then tan θ/2 = (1-cos θ)/(sin θ) = (2/3)/(2√2/3) = 1/√2, so the second term is 4√2 × 1/√2 = 4.
4. So. The expression equals 25 + 4 = 29.

The common slip: 17. Solving 2t/(1-t²) = 2√2 and taking the negative root t = -√2 makes the second term -8 and the total 17. Since θ is acute, θ/2 is acute and tan θ/2 must be positive.

The idea to keep. Check whether two inverse-trig terms are complements before you expand anything. For a half angle, tan θ/2 = (1-cos θ)/(sin θ) gives the sign directly.

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