JEE Main 2021 · 26 August · morning · Single correct · +4 / −1
The sum of solutions of the equation
(cos x)/(1 + sin x) = | tan 2x |, x ∈ ( - π/2,π/2 ) - { π/4, - π/4 } is :
- (a) - 11π/30
- (b) π/10
- (c) - 7π/30
- (d) - π/15
Answer and solution
Answer (a) - 11π/30
Solution
Maths sir · 4 steps
Turn the left side into tan(π/4-x/2), then open the modulus in two cases.
- Rewrite the left side. Write cos x = cos² x/2-sin² x/2 and 1+sin x = (cos x/2+sin x/2)², cancel the common factor and divide top and bottom by cos x/2. This leaves (1-tan x/2)/(1+tan x/2) = tan(π/4-x/2). For x ∈ (-π/2,π/2) this angle lies in (0,π/2), so the left side is always positive.
tan(π/4-x/2) = |tan 2x|
- Positive case. If tan 2x > 0, then 2x = π/4-x/2+nπ, which gives x = π/10+2nπ/5. Inside the interval, n = 0 gives π/10 and n = -1 gives -3π/10. For both, tan 2x > 0 holds.
- Negative case. If tan 2x < 0, then -2x = π/4-x/2+nπ, which gives x = -π/6-2nπ/3. Only n = 0 lands inside the interval, giving -π/6. There tan 2x = tan(-π/3) < 0, as needed.
- So. The sum is π/10-3π/10-π/6 = -π/5-π/6 = -11π/30, which is option (a).
The idea to keep. (cos x)/(1+sin x) is always tan(π/4-x/2). With a modulus, split into sign cases and list every n that falls inside the interval, not just n = 0.
Answer from the NTA final answer key. Solution by PrepWiser.