Question

Set $\frac{tanA+tanB}{1-tanAtanB} = \frac{3}{\sqrt{3} }$ when 0 < A < 90° and 0 < B < 90° find the value of A+B

Answer 1

Answer:

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Answer 2

Answer:

A + B = 90° => A = 90 - B

So Tan A = Cot (90 - A) = Cot B

So Tan B = Cot (90 - B) = Cot A

SecB = Cosec (90 -B) = Cosec A

CosA = Sin (90 -A) = Sin B

substitute these in the LHS,

TanA\ TanB+\frac{TanA\ CotB}{SinA \ SecB}-\frac{Sin^2B}{Cos^2A}\\\\=TanA\ CotA + \frac{TanA\ TanA}{SinA\ CosecA}-\frac{Sin^2B}{Sin^2B}\\\\=1+Tan^2A - 1=Tan^2A

Step-by-step explanation:

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MARK AS BRANILY