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If 2 tan A+cot A = tan B then cot A+2tan(A- B) =A)-1B) 0C) 1D) 1/2 |
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Answer» 2TanA+CotA=TanBwe will substitute this value of TanB while solving the answer.CotA + 2Tan(A-B)= CotA + 2 [ (TanA-TanB)/ (1+TanATanB)]= CotA+ 2 [(TanA-2TanA-CotA)/ (1+TanA(2Tan+CotA))]= CotA + 2 [(-TanA-CotA)/(1+2Tan2A+1)]= CotA + 2 [(-TanA-CotA)/(2+2Tan2A)]=CotA+ [(-TanA-CotA)/(1+Tan2A)]= CotA+ [(-TanA-CotA)/(Sec2A)]= CotA - TanA/Sec2A - CotA/Sec2A = CotA - SinACosA - CotACos2A= CotA(1-Cos2A) - SinACosA= CotA(Sin2A) - SinACosA= SinACosA - SinACosA= 0therefore CotA + 2Tan(A-B) = 0 |
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