Question #75f9a

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Answer 1 · 2016-11-23T16:28:31

Given #tan(alpha-beta)=7/24 and tanalpha=4/3#

So #(tanalpha-tanbeta)/(1+tanalphatanbeta)=7/24#

#=>(4/3-tanbeta)/(1+4/3tanbeta)=7/24#

#=>32-24tanbeta=7+28/3tanbeta#

#=>100/3tanbeta=25#

#=>tanbeta=3/4#

So #cotalpha =3/4 and cotbeta=4/3#

Now

#cot(alpha+beta)=(cotalphacotbeta-1)/(cotbeta+cotalpha)#

#=>cot(alpha+beta)=(3/4xx4/3-1)/(4/3+3/4)=0=cot(pi/2)#

#=>alpha+beta=pi/2#

proved