Question

If tan α = 1/7 , tan β = 1/3, then cos 2α is equal to
(A) sin 2β (B) sin 4β (C) sin 3β (D) cos 2β

Answer 1

   2 * tan-1 {tan α/2 * tan(π/4 - β/2)}

= tan-1 [{2 * tan α/2 * tan π/4 - β/2)}/{1 - tan2 α/2 * tan2 (π/4 - β/2)}]

= tan-1 [{2 * tan α/2 * (1 - tan β/2)/(1 + tan β/2)}/{1 - tan2 α/2 * (1 - tan β/2)2 /((1 + tan β/2)2 }]

= tan-1 [{2 * tan α/2 * (1 - tan β/2) * (1 + tan β/2)}/{(1 + tan β/2)2 - tan2 α/2 * (1 - tan β/2)2 }]

= tan-1 [{2 * tan α/2 * (1 - tan2 β/2)}/{1 + tan2 β/2 + 2*tan β/2 - tan2 α/2 * (1 + tan2 β/2 - 2*tan

β/2}]

= tan-1 [{2 * tan α/2 * (1 - tan2 β/2)}/{(1 + tan2 β/2)* (1 - tan2 α/2)  + 2 *tan β/2(1 + tan2 α/2}]

= tan-1 [{2*tan α/2)/(1 + tan2 α/2)}*{1 - tan2 β/2)/(1 + tan2 β/2)} /{(1 - tan2 α/2)/(1 + tan2 α/2)  + 2

*tan β/2(1 + tan2 β/2}] 

= tan-1 {(sin α * cos β)/(cos α + cos β)}

[used different inverse trigometory formulas]