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Question
Show that `2tan^-1 {tan alpha/2 * tan(pi/4 - beta/2)} = tan^-1 (sin alpha cos beta)/(cosalpha + sinbeta)`
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Solution
L.H.S. = `tan^-1 (2tan alpha/2 * tan (pi/4 - beta/2))/(1 - tan^2 alpha/2 tan^2 (pi/4 - beta/2))` ......`("since" 2 tan^-1x = tan^-1 (2x)/(1 - x^2))`
= `tan^-1 (2tan alpha/2 (1 - tan beta/2)/(1 + tan beta/2))/(1 - tan^2 alpha/2 ((1 - tan beta/2)/(1 + tan beta/2))^2)`
= `tan^-1 (2tan alpha/2 (1 - tan^2 beta/2))/((1 + tan beta/2)^2 - tan^2 alpha/2 (1 - tan beta/2)^2)`
= `tan^-1 (2tan alpha/2 (1 - tan^2 beta/2))/((1 + tan^2 beta/2)(1 - tan^2 alpha/2) + 2 beta/2 (1 + tan^2 alpha/2))`
= `tan^-1 ((2tan alpha/2)/(1 + tan^2 alpha/2) - (1 - tan^2 beta/2)/(1 + tan^2 beta/2))/((1 - tan^2 alpha/2)/(1 + tan^2 alpha/2) + (2tan beta/2)/(1 + tan^ beta/2))`
= `tan^-1 ((sin alpha cos beta)/(cos alpha + sin beta))`
= R.H.S.
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