tan (2 tan ⁻¹ 1/5+sec ⁻¹ √5/2+2 tan ⁻¹ 1/8) is equal to :

Solution

Maths sir · 4 steps

Group the two doubled terms first, add the tangents, then turn the secant into an inverse tangent.

1. Pair the doubles. Take out the common 2: 2(tan⁻¹ 1/5+tan⁻¹ 1/8). Since 1/5×1/8<1, the bracket is tan⁻¹(1/5+1/8)/(1-1/40) = tan⁻¹(13/40)/(39/40) = tan⁻¹ 1/3.
2. Now double it. Use 2 tan⁻¹ x = tan⁻¹ 2x/(1-x²) for |x|<1. With x=1/3 this gives tan⁻¹(2/3)/(8/9) = tan⁻¹ 3/4.
3. Convert the secant. If sec θ = √5/2 with θ acute, then cos θ = 2/√5 and sin θ = 1/√5. So tan θ = 1/2, that is sec⁻¹√5/2 = tan⁻¹ 1/2.

   tan⁻¹ 3/4+tan⁻¹ 1/2 = tan⁻¹(3/4+1/2)/(1-3/8) = tan⁻¹(5/4)/(5/8) = tan⁻¹ 2

4. So. The whole angle is tan⁻¹ 2, so its tangent is 2, option (b).

The idea to keep. Before using any formula, write every inverse function as tan⁻¹ of something. Then the sum formula and 2 tan⁻¹ x = tan⁻¹ 2x/(1-x²) do all the work.

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