JEE Main 2021 · 26 February · morning · Single correct · +4 / −1
If (sin ¹x)/a = (cos ⁻¹x)/b = (tan ⁻¹y)/c; 0 < x < 1,
then the value of cos ( (π c)/(a + b) ) is :
- (a) (1 - y²)/2y
- (b) (1 - y²)/y√y
- (c) 1 - y²
- (d) (1 - y²)/(1 + y²)
Answer and solution
Answer (d) (1 - y²)/(1 + y²)
Solution
Maths sir · 4 steps
Call the common ratio k. Adding the first two angles gives π/2, which fixes k at once.
- Name the ratio. Let each ratio equal k. Then sin⁻¹ x = ak, cos⁻¹ x = bk and tan⁻¹ y = ck.
- Find k. Use sin⁻¹ x + cos⁻¹ x = π/2. This gives (a+b)k = π/2, so k = π/2(a+b).
- Rewrite the angle. Now (π c)/(a+b) = 2ck = 2 tan⁻¹ y. Put θ = tan⁻¹ y, so tan θ = y and you need cos 2θ.
cos 2θ = (1-tan² θ)/(1+tan² θ) = (1-y²)/(1+y²)
- So. cos((π c)/(a+b)) = (1-y²)/(1+y²), which is option (d).
The idea to keep. When inverse trig ratios are set equal, call the common value k. Then look for a pair like sin⁻¹ x + cos⁻¹ x = π/2 that pins k down.
Answer from the NTA final answer key. Solution by PrepWiser.