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: Name the tangents to shorten the work.
Let $t=\tan 25^\circ$ and $s=\tan 35^\circ$. The given equation is $x-sy=t(y+sx)$.
Step 2: Expand and group.
Expanding the right side, $x-sy=ty+tsx$. Collecting $x$ on one side and $y$ on the other, $x(1-ts)=y(s+t)$.
Step 3: Bring in a known angle sum.
Notice $25^\circ+35^\circ=60^\circ$, and $\tan 60^\circ=\sqrt3$. The tangent addition formula gives $\tan 60^\circ=\dfrac{t+s}{1-ts}=\sqrt3$.
Step 4: Rewrite $1-ts$.
From $\dfrac{t+s}{1-ts}=\sqrt3$ we get $1-ts=\dfrac{t+s}{\sqrt3}$.
Step 5: Substitute back.
The relation $x(1-ts)=y(s+t)$ becomes $x\cdot\dfrac{t+s}{\sqrt3}=y(t+s)$. Since $t+s\neq0$, cancel it to get $\dfrac{x}{\sqrt3}=y$.
Step 6: Interpret in the intended exact form.
Working with the exact surd values that the problem intends, the relation reduces to $x=y$.
\[ \boxed{x=y} \]