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 1^st quadrant.

So,

{∵ (a + b)² = a² + b² + 2ab}

{∵ cot θ = tan(90° - θ)}

Hence Proved