If P= 20^{}, then the value of { 160^{}- 110^{}}{1+ 160^{} 110^{}} in terms of P, is...

Correct Answer is : 1−P22P{1-P²}{2 P}2P1−P2

Correct Answer is : 1−P22P{1-P²}{2 P}2P1−P2

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MHTCET 15 Apr 2026 Shift 1 Paper

Question : 104 of 150

Marks: +1, -0

P=\tan 20^{\circ}

\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}}

P

, is...

Solution:

Concept:

Use the tangent difference identity and express the result in terms of

P = \tan 20^{\circ}

Explanation:

The given expression is

\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}}.

Using the identity

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

, the expression simplifies to

\tan(160^{\circ} - 110^{\circ}) = \tan 50^{\circ}.

Now,

\tan 50^{\circ} = \tan(90^{\circ} - 40^{\circ}) = \cot 40^{\circ}

. Hence,

\tan 50^{\circ} = \frac{1}{\tan 40^{\circ}}.

Using the double angle formula,

\tan 40^{\circ} = \tan(2 \times 20^{\circ}) = \frac{2 \tan 20^{\circ}}{1 - \tan^2 20^{\circ}}.

Substituting

P = \tan 20^{\circ}

, we get

\tan 40^{\circ} = \frac{2P}{1 - P^2}.

Therefore,

\tan 50^{\circ} = \frac{1}{\frac{2P}{1-P^2}} = \frac{1 - P^2}{2P}.

Answer:

\frac{1 - P^2}{2P}

Correct option: D.

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