Evaluate tan{(1/2)[sin⁻¹(2x/(1+x²))+cos⁻¹((1−y²)/(1+y²))]}, |x| 0, xy<1.

Final Answer: (x+y)/(1−xy) Step 1: Let α=tan⁻¹x∈(−π/4,π/4), β=tan⁻¹y∈(0,π/2). Step 2: sin 2α=2x/(1+x²), with 2α in the sine-inverse principal range. Likewise cos 2β=(1−y²)/(1+y²), with 2β∈(0,π). Step 3: Therefore the expression is tan(α+β). The condition xy 0, so its denominator is nonzero. Step 4: Use tan(α+β)=(tan α+tan β)/(1−tan α tan β)=(x+y)/(1−xy).

NCERT Class 12 · Ch 2 Exercise 2.2 · Q9

NCERT Class 12 Maths · Chapter 2 Inverse Trigonometric Functions · Exercise 2.2 Question 9

Evaluate tan{(1/2)[sin⁻¹(2x/(1+x²))+cos⁻¹((1−y²)/(1+y²))]}, |x|<1, y>0, xy<1.

💡 Concept Hints

Recall the principal range of the inverse function before simplifying.
Find an angle with the required trigonometric value, then verify it belongs to the principal range.

✅ Step-by-Step Solution

Final Answer: (x+y)/(1−xy)

Let α=tan⁻¹x∈(−π/4,π/4), β=tan⁻¹y∈(0,π/2).
sin 2α=2x/(1+x²), with 2α in the sine-inverse principal range. Likewise cos 2β=(1−y²)/(1+y²), with 2β∈(0,π).
Therefore the expression is tan(α+β). The condition xy<1 ensures 1−xy>0, so its denominator is nonzero.
Use tan(α+β)=(tan α+tan β)/(1−tan α tan β)=(x+y)/(1−xy).