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NCERT Exemplar · Q38

Q.The value of tan⁡75∘−cot⁡75∘\tan 75^\circ - \cot 75^\circ is equal to
(A) 232\sqrt{3}
(B) 2+32 + \sqrt{3}
(C) 2−32 - \sqrt{3}
(D) 11

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The key idea is to rewrite cot⁡75∘\cot 75^\circ as tan⁡(90∘−75∘)=tan⁡15∘\tan(90^\circ - 75^\circ) = \tan 15^\circ, then use the tangent subtraction formula to evaluate tan⁡75∘−tan⁡15∘\tan 75^\circ - \tan 15^\circ. The final value is 232\sqrt{3}, which corresponds to option (A).


Concept and Intuition

This problem tests your comfort with complementary angle identities and the tangent subtraction formula. The trap is that tan⁡75∘\tan 75^\circ and cot⁡75∘\cot 75^\circ are not directly combinable — but cot⁡θ=tan⁡(90∘−θ)\cot \theta = \tan(90^\circ - \theta) turns the expression into a difference of two tangents: tan⁡75∘−tan⁡15∘\tan 75^\circ - \tan 15^\circ. That difference has a neat closed form.

Watch out

A common mistake is to compute tan⁡75∘\tan 75^\circ and cot⁡75∘\cot 75^\circ separately using tan⁡75∘=2+3\tan 75^\circ = 2 + \sqrt{3} and cot⁡75∘=2−3\cot 75^\circ = 2 - \sqrt{3}, then subtract. That works, but it’s slower and risks sign errors. The complementary-angle trick is faster and cleaner.


Step-by-Step Solution

1. Rewrite cot⁡75∘\cot 75^\circ using the complementary angle identity.

We know that for any angle θ\theta,

cot⁡θ=tan⁡(90∘−θ).\cot \theta = \tan(90^\circ - \theta).

So

cot⁡75∘=tan⁡(90∘−75∘)=tan⁡15∘.\cot 75^\circ = \tan(90^\circ - 75^\circ) = \tan 15^\circ.

Thus the given expression becomes:

tan⁡75∘−cot⁡75∘=tan⁡75∘−tan⁡15∘.\tan 75^\circ - \cot 75^\circ = \tan 75^\circ - \tan 15^\circ.

2. Recall the tangent subtraction formula.

For any two angles AA and BB,

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B.\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.

This formula lets us express tan⁡A−tan⁡B\tan A - \tan B in terms of tan⁡(A−B)\tan(A-B) and the product tan⁡Atan⁡B\tan A \tan B:

tan⁡A−tan⁡B=tan⁡(A−B)⋅(1+tan⁡Atan⁡B).\tan A - \tan B = \tan(A - B) \cdot (1 + \tan A \tan B).

3. Apply the formula with A=75∘A = 75^\circ and B=15∘B = 15^\circ.

Here A−B=60∘A - B = 60^\circ, and tan⁡60∘=3\tan 60^\circ = \sqrt{3}. So …

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