Prove that: 
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Proof:
Identities used:
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Therefore,
tan 15° = tan (45° - 30°)
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On rationalising:


{∵ (a – b)(a + b) = a2 – b2}
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On rationalising


{∵ (a – b)(a + b) = a2 – b2}
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Let 2θ = 15°
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We know,
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Formula used:
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{∵ (a + b)2 = a2 + b2 + 2ab}

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cot θ < 0 as θ is in 1st quadrant.
So,
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{∵ (a + b)2 = a2 + b2 + 2ab}
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{∵ cot θ = tan(90° - θ)}
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Hence Proved
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