Find the values of the following: tan(1/2)[sin⁻¹(2x/(1+x²)) + cos⁻¹((1-y²)/(1+y²))], |x| < 1, y > 0 and xy < 1
The value is (x + y)/(1 - xy).
Let x = tan α and y = tan β.
Step 1: Simplify sin⁻¹(2x/(1+x²))
Since |x| < 1, we have |tan α| < 1, so -π/4 < α < π/4. Now, 2x/(1+x²) = 2tan α/(1+tan²α) = 2tan α/sec²α = 2sin α cos α = sin 2α
Therefore, sin⁻¹(2x/(1+x²)) = sin⁻¹(sin 2α) = 2α = 2tan⁻¹x
Step 2: Simplify cos⁻¹((1-y²)/(1+y²))
Since y > 0, we have tan β > 0, so 0 < β < π/2. Now, (1-y²)/(1+y²) = (1-tan²β)/(1+tan²β) = (1-tan²β)/sec²β = cos²β - sin²β = cos 2β
Therefore, cos⁻¹((1-y²)/(1+y²)) = cos⁻¹(cos 2β) = 2β = 2tan⁻¹y
Step 3: Find the final value
tan(1/2)[2tan⁻¹x + 2tan⁻¹y] = tan[tan⁻¹x + tan⁻¹y]
Using tan(A+B) = (tan A + tan B)/(1 - tan A tan B):
= (x + y)/(1 - xy), which is valid since xy < 1.
Explanation
This question tests the standard substitutions for inverse trigonometric expressions. The key insight is recognizing that sin⁻¹(2x/(1+x²)) = 2tan⁻¹x when |x| < 1, and cos⁻¹((1-y²)/(1+y²)) = 2tan⁻¹y when y > 0. These are derived using the substitutions x = tan α and y = tan β, along with the double angle formulas sin 2α = 2sin α cos α and cos 2β = cos²β - sin²β. The final step uses the tangent addition formula.
Solution Steps
Step 1: Substitute x = tan α and simplify sin⁻¹(2x/(1+x²)) to get 2tan⁻¹x
Step 2: Substitute y = tan β and simplify cos⁻¹((1-y²)/(1+y²)) to get 2tan⁻¹y
Step 3: Apply tan to the sum and use tan(A+B) formula to get (x+y)/(1-xy)