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: Understanding the Concept:
The expression has the form \(\dfrac{\tan x-\tan y}{1+\tan x\tan y}=\tan(x-y)\).

Step 2: Apply:
Here \(x=160^\circ\) and \(y=110^\circ\). So the value is \(\tan(160^\circ-110^\circ)=\tan 50^\circ\).

Step 3: Write in terms of tan 20:
Use \(\tan 50^\circ=\tan(90^\circ-40^\circ)=\cot 40^\circ=\dfrac{1}{\tan 40^\circ}\). With \(\tan 40^\circ=\dfrac{2P}{1-P^2}\) (double angle for \(P=\tan 20^\circ\)),
\[ \tan 50^\circ=\frac{1-P^2}{2P} \]

Step 4: Choose:
This is option (D). Option (B) has \(1+P^2\) and would be wrong because the double angle formula has \(1-P^2\) in the denominator of \(\tan 40^\circ\).

Final Answer:
The value is tan 50 degrees = (1-P^2)/(2P). \[ \boxed{\frac{1-P^2}{2P}} \]