Find exact value of tan(t+x) if sin(t)=3/5 and sin(x)=12/13 and t and x are between 0 pi/2?

Question

between 0 and pi/2

Help much appreciated!

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Answer 1 · 2018-05-19T17:32:35

#-3.94# or #-63/16#

So, by formula,
#tan(t+x)=(tant +tanx)/(1-tant*tanx)#

Proof
#tan(t+x)=sin(t+x)/cos(t+x)#

#=(sint*cosx+cost*sinx)/(cost*cosx-sint*sinx)#

divide Numerator and Denominator by #cost*cosx # and separate the terms, and you"ll get,
#tan(t+x)=(sint/cost +sinx/cos)/(1-sint/cost*sinx/cosx)#

#=(tant +tanx)/(1-tant*tanx)#

Coming back to the problem,
as you know #sint=3/5#,
#tant = sint/cost#

#=sint/sqrt(1-sin^2t)#

#=3/5/4/5 = 3/4#

Similarly,
#tanx = sinx/cosx#

#=sinx/sqrt(1-sin^2x)#

#=12/13/5/13 = 12/5#

From the above 2,
#tan(t+x)= (3/4+12/5)/(1-3/4*12/5)#

#=63/-16#

#=-3.94#

Hope this Helps! :)