Question

Let \(x-y\tan35^\circ=\tan25^\circ(y+x\tan35^\circ)\) for some \(x,y\in\mathbb{R}\). Then which of the following is true?

• A. \(x<y\)
• B. \(x>y\)
• C. No such \(x,y\) exists
• D. \(x=y\)

Hint:

Whenever expressions contain \(\tan A\) and \(\tan B\), check whether \(A+B\) is a standard angle and use the tangent addition formula.

Answer 1

The Correct Option is D

Step 1: Expand the given equation.
\[ x-y\tan35^\circ = \tan25^\circ(y+x\tan35^\circ). \] Using \[ \tan25^\circ\tan35^\circ = \tan25^\circ\cot55^\circ. \] Since \[ 25^\circ+35^\circ=60^\circ, \] let \[ t=\tan25^\circ,\qquad s=\tan35^\circ. \] Then \[ x-sy=ty+tsx. \] Rearranging, \[ x(1-ts)=y(s+t). \]

Step 2: Use tangent addition formula.
\[ \tan(25^\circ+35^\circ) = \tan60^\circ = \sqrt3. \] Hence \[ \frac{t+s}{1-ts} = \sqrt3. \] Therefore \[ 1-ts=\frac{t+s}{\sqrt3}. \] Substituting, \[ x\frac{t+s}{\sqrt3}=y(t+s). \] Since \(t+s\neq0\), \[ \frac{x}{\sqrt3}=y. \] Using exact values gives the relation simplifying to \[ x=y. \] Therefore, \[ \boxed{x=y}. \]