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[SOLVED] Trig Identities and Double Angle Formulas
If tan(a) = 1/5 and tan(b) = 1/239, find tan(4a - b)
tan2x = 2tanx/1-tan^2x
tan(x-y) = tanx - tany/ 1 +tanx tany
tan (4a - b) = (tan4a - tanb)/(1 + tan4a tanb)
= ((2tan2a/1-tan^2 2a) - tanb)/ (1-(2tan^2 2a/1-tan^2 2a)tanb)
= ((2tan2(1/5)/1-tan^2 2(1/5)) - tan(1/239)) / (1 - (2tan^2 2(1/5))tan(1/239))
= ((2tan(2/5))/1-tan^2 (2/5)) - tan(1/239)) / (1 - 2tan^2 (2/5)tan(1/239))
After this I get stuck. Please someone help. I need this really quick.
Homework Statement
If tan(a) = 1/5 and tan(b) = 1/239, find tan(4a - b)
Homework Equations
tan2x = 2tanx/1-tan^2x
tan(x-y) = tanx - tany/ 1 +tanx tany
The Attempt at a Solution
tan (4a - b) = (tan4a - tanb)/(1 + tan4a tanb)
= ((2tan2a/1-tan^2 2a) - tanb)/ (1-(2tan^2 2a/1-tan^2 2a)tanb)
= ((2tan2(1/5)/1-tan^2 2(1/5)) - tan(1/239)) / (1 - (2tan^2 2(1/5))tan(1/239))
= ((2tan(2/5))/1-tan^2 (2/5)) - tan(1/239)) / (1 - 2tan^2 (2/5)tan(1/239))
After this I get stuck. Please someone help. I need this really quick.