if tan(α)=1/3 and tan(x)=1/7 then 2α+x
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нεяε ¡ร тнε คหรωεя
тคห x = 2/3 คหd тคห ყ = 3/4.
หσω, тคห (x+ყ) = [тคห x + тคห ყ]/[1 - тคห x*тคห ყ]
= [(2/3) + (3/4)]/[1 - (2/3)*(3/4)]
= [8+9]/12*[1 - (2/3)*(3/4)]
= 17/12*(1/2)
= 17/6
คหรωεя: тคห (x+ყ) = 17/6.
нσpε ¡т нεłpร
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adikarthikeya:
Tan*-1 ( 11/2)
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