Proof:

Identities used:

Therefore,

tan 15° = tan (45° - 30°)

On rationalising:

{ (a – b)(a + b) = a2 – b2}

On rationalising

{ (a – b)(a + b) = a2 – b2}

Let 2θ = 15°

We know,

Formula used:

{ (a + b)2 = a2 + b2 + 2ab}

cot θ < 0 as θ is in 1st quadrant.

So,

{ (a + b)2 = a2 + b2 + 2ab}

{ cot θ = tan(90° - θ)}

Hence Proved

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