Q. 386. Trigonometric identities.

Question 386.
Prove that
1) tan20° tan40° tan80° = √3
2) tan(a-b)/2 = √{(4-x^2-y^2)/(x^2 + y^2)} where x = sin a + sin b & y = cos a + cos b.

Answer 386.
1)
tanθ tan(60° - θ) tan (60° + θ)
=  tanθ [(tan60° - tanθ) / (1 + tan60° tanθ)] [(tan60° + tanθ) / (1 - tan60° tanθ)]
=  tanθ [(√3 - tanθ) / (1 + √3 tanθ)] [(√3 + tanθ) / (1 - √3 tanθ)]
= tanθ (3 - tan^2 θ) / (1 - √3 tan^2 θ)
= (3tanθ - tan^3 θ) / (1 - 3tan^2 θ)
= tan3θ.

Taking θ = 20°,
tan20° tan40° tan80°
= tan60°
= √3.

 2)
x = sina + sinb and y = cosa + cosb
Squarring and adding the given eqns.,
(sina + sinb)^2 + (cosa + cosb)^2 = x^2 + y^2
=> (sin^2 a + cos^2 a) + (sin^2 b + cos^2 b) + 2 (cosa cosb + sina sinb) = x^2 + y^2
=> 2 [1 + cos(a - b)] = x^2 + y^2
=> cos(a - b) = (x^2 + y^2)/2 - 1
=> [1 - tan^2 (a-b)/2] / [1 + tan^2 (a-b)/2] = (x^2 + y^2 - 2)/2
=> [1 + tan^2 (a-b)/2 - 1 + tan^2 (a-b)/2] / [1 + tan^2 (a-b)/2 + 1 - tan^2 (a-b)/2]
......................= (2 - x^2 - y^2 + 2) / (2 + x^2 + y^2 - 2)
=> tan^2 (a-b)/2 = (4 - x^2 - y^2) / (x^2 + y^2)
=> tan(a - b)/2 = √ [(4 - x^2 - y^2) / (x^2 + y^2)].

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