tanA = m/m+1 and tanB = 1/2m+1 prove that tanA + tanB + tanAtanB = 1
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tanA + tanB + tanAtanB
m/(m+1) + 1/(2m+1) + m1/(m+1)(2m+1)
[m(2m+1)+1(m+1)+m] / [(m+1)(2m+1)]
[m(2m+1)+m+(m+1)] / [(m+1)(2m+1)]
[m(2m+1+1)+(m+1)] / [(m+1)(2m+1)]
[m(2m+2)+(m+1)] / [(m+1)(2m+1)]
[2m(m+1)+(m+1)] / [(m+1)(2m+1)]
[(2m+1)(m+1)] / [(m+1)(2m+1)]
= 1
Or see this:
m/(m+1) + 1/(2m+1) + m1/(m+1)(2m+1)
[m(2m+1)+1(m+1)+m] / [(m+1)(2m+1)]
[m(2m+1)+m+(m+1)] / [(m+1)(2m+1)]
[m(2m+1+1)+(m+1)] / [(m+1)(2m+1)]
[m(2m+2)+(m+1)] / [(m+1)(2m+1)]
[2m(m+1)+(m+1)] / [(m+1)(2m+1)]
[(2m+1)(m+1)] / [(m+1)(2m+1)]
= 1
Or see this:
Attachments:
Arnav987:
thanks
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