Prove that 1 - tan- tan-2 - -

LHS =1+tanα·tanα2=1+sinα·sinα/2cosα·cosα/2 =cosα·cosα/2+sinα·sinα/2cosα·cosα/2 =cos(α α/2)cosα·cosα/2=cosα/2cosα·cosα/2 =1cosα=α= RHS. ...

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Cos(A+B)Cos(A-B)

Prove that ...

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Prove that $1 + tan α \cdot t a n \frac{α}{2} = sec α$

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(tan α {sec}^{2} α + cot α c o {sec}^{2} α) - (tan α + cot α) = 2 c o sec 2 α [{sec}^{2} α + c o {sec}^{2} α - 3]

1 + \frac{1}{2} cos α + \frac{1.3}{2.4} cos 2 α + \frac{1.3.5}{2.4.6} cos 3 α + \dots = {(2 sin \frac{α}{2})}^{- 1 / 2} [cos \frac{π - α}{4} + i sin \frac{π - α}{4}]

α + β = π / 4

α

β

α + β = \frac{π}{4}

(1 + t a n α) (1 + t a n β) = 2

cos θ = cos α cos β

tan \frac{θ + α}{2} tan \frac{θ - α}{2} = {tan}^{2} \frac{β}{2}

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