Question

if tan B=n sinA cosA/1-n (sinA)^2 then show that tan(A+B) =(1-n) tanA​

Answer 1

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tanb= nsina cosa/1 - n sin^2a

tan( a- b)= tana - tanb/1+ tana tanb

={ tana - (nsiacosa/1-nsina^2a ]/1+ tana ( nsina cosa/1- nsin^2a)

= [tana*(1- nsin^2a) - nsina cosa]/ {1- nsin^2a +tana(nsina cosa }

=tana - nsin^2a *tana - nsina cosa]/1- nsin^2a + n sin^2a

= tana[ 1- nsin^2 - ncos^2a]/1

= tana[ 1- n(sin^2a + cos^2a) ]

= tana ( 1- n)

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