ax² + bx + c = 0 সমীকরণের ক্ষেত্রে 9ac = 2b² হলে তার মূলদ্বয়ের ধরণ হবে-

2 tan⁻¹(1/3) + tan⁻¹(1/7) = tan⁻¹(2*(1/3) / (1 - (1/3)²)) + tan⁻¹(1/7) = tan⁻¹((2/3) / (8/9)) + tan⁻¹(1/7) = tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹((3/4 + 1/7) / (1 - (3/4)*(1/7))) = tan⁻¹((21+4)/28 / (28-3)/28) = tan⁻¹(25/25) = tan⁻¹(1) = π/4.

This is a standard trigonometric identity: cosec⁻¹x + sec⁻¹x = π/2.

cosec²(tan⁻¹ 1/2) - 3 sec²(cot⁻¹ sqrt(3)) = (1 + cot²(tan⁻¹ 1/2)) - 3(1 + tan²(cot⁻¹ sqrt(3))) = (1 + (1/(1/2))²) - 3(1 + (1/sqrt(3))²) = (1 + 2²) - 3(1 + 1/3) = (1 + 4) - 3(4/3) = 5 - 4 = 1.

tan⁻¹(1/3) + tan⁻¹(1/7) = tan⁻¹((1/3 + 1/7)/(1 - 1/3 × 1/7)) = tan⁻¹((10/21)/(20/21)) = tan⁻¹(1/2).

sin(2 sin⁻¹(1/2)) = sin(2 * π/6) = sin(π/3) = √3/2.

sin⁻¹(2/sqrt(5)) = tan⁻¹(2/1) = tan⁻¹(2). The options provided in the image seem to have a typo in the question or options. Based on the triangle with perpendicular 2 and hypotenuse sqrt(5), the base is sqrt((sqrt(5))² - 2²) = sqrt(5-4) = 1. So, sin⁻¹(2/sqrt(5)) = tan⁻¹(2/1) = tan⁻¹(2). Looking at the provided solution in the image: sin⁻¹(2/sqrt(5)) = tan⁻¹(2/1) = cot⁻¹(1/2). The image shows the answer as tan⁻¹(1/sqrt(5)) is not correct based on the triangle. Wait, the image shows: sin⁻¹(2/sqrt(5)) = tan⁻¹(2/1) = cot⁻¹(1/2). Actually, the image shows sin⁻¹(2/sqrt(5)) = tan⁻¹(2/1) = cot⁻¹(1/2). Let's re-read the image: sin⁻¹(2/sqrt(5)) = tan⁻¹(2/1) = cot⁻¹(1/2). The option D is tan⁻¹(1/sqrt(5)). There might be a typo in the question or options in the source.

Let tan⁻¹(4/3) = θ, then tanθ = 4/3. In a right triangle, opposite=4, adjacent=3, hypotenuse=5. sinθ = 4/5.

Let sin⁻¹x = θ, then x = sinθ. √(1-x²) = cosθ. tan⁻¹(cosθ) = tan⁻¹(sin(π/2-θ)/cos(π/2-θ)) = π/2-θ = cos⁻¹x.

cos⁻¹(sinx) = cos⁻¹(cos(π/2 - x)) = π/2 - x.

1/2 sin⁻¹(2x/(1+x²)) + cot⁻¹(1/x) = 1/2 × 2tan⁻¹x + tan⁻¹x = tan⁻¹x + tan⁻¹x = 2tan⁻¹x + cot⁻¹x = tan⁻¹x + (tan⁻¹x + cot⁻¹x) = tan⁻¹x + π/2. Given the options, the simplified form is π.