Question

If \(P = tan20^{\circ}\), then the value of \(\frac{tan160^{\circ}-tan110^{\circ}}{1+tan160^{\circ}tan110^{\circ}}\) in terms of \(P\), is...

• A. \(\frac{1+P^2}{2P^2}\)
• B. \(\frac{1+P^2}{2P}\)
• C. \(\frac{1-P^2}{2P^2}\)
• D. \(\frac{1-P^2}{2P}\)

Hint:

Use tan(x - y) formula, or reduce each tangent to P.

Answer 1

The Correct Option is D

Step 1: Reduce each tangent:
$\tan160^\circ=-\tan20^\circ=-P$. $\tan110^\circ=-\cot20^\circ=-1/P$.

Step 2: Substitute:
Numerator: $-P+\frac1P=\frac{1-P^2}{P}$. Denominator: $1+(-P)(-1/P)=2$.

Step 3: Divide:
\[ \frac{(1-P^2)/P}{2}=\frac{1-P^2}{2P} \]

Final Answer:
Direct substitution gives (1-P^2)/(2P). \[ \boxed{D} \]