Question

Let \(m\) and \(n\) be non–negative integers such that for \[ x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\qquad \tan x+\sin x=m,\quad \tan x-\sin x=n. \] Then the possible ordered pair \((m,n)\) is:

• A. \((2,1)\) but not \((3,4)\)
• B. \((3,4)\) but not \((2,1)\)
• C. both \((2,1)\) and \((3,4)\)
• D. neither \((2,1)\) nor \((3,4)\)

Hint:

When trigonometric expressions are given as sums and differences, always reduce them to \(\sin x\) and \(\tan x\), then use \(\sin^2 x+\cos^2 x=1\) to check consistency.

Answer 1

The Correct Option is D

Step 1: Add and subtract the given equations \[ \tan x+\sin x=m,\qquad \tan x-\sin x=n \] Adding, \[ 2\tan x=m+n \Rightarrow \tan x=\frac{m+n}{2} \] Subtracting, \[ 2\sin x=m-n \Rightarrow \sin x=\frac{m-n}{2} \] Step 2: Use the identity \(\tan x=\dfrac{\sin x}{\cos x}\) \[ \cos x=\frac{\sin x}{\tan x} =\frac{(m-n)/2}{(m+n)/2} =\frac{m-n}{m+n} \] Step 3: Apply the identity \(\sin^2 x+\cos^2 x=1\) \[ \left(\frac{m-n}{2}\right)^2+\left(\frac{m-n}{m+n}\right)^2=1 \] \[ \frac{(m-n)^2}{4}+\frac{(m-n)^2}{(m+n)^2}=1 \] Step 4: Test the given pairs Case (i): \((m,n)=(2,1)\) \[ m-n=1,\quad m+n=3 \] \[ \frac{1}{4}+\frac{1}{9}=\frac{13}{36}\neq 1 \] So \((2,1)\) is not possible. Case (ii): \((m,n)=(3,4)\) \[ m-n=-1,\quad m+n=7 \] \[ \frac{1}{4}+\frac{1}{49}=\frac{53}{196}\neq 1 \] So \((3,4)\) is also not possible. Final Conclusion: Neither \((2,1)\) nor \((3,4)\) satisfies the required condition. \[ \boxed{\text{Option (D)}} \]