If p = 2sinA /1+cosA+sinA and q = cosA / 1+sinA ,Then prove that p+q = 1

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Krishna Kumar

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Given that p = 2sinA /1+cosA+sinA and q = cosA / 1+sinA We have to prove p+q = 1 Proof: LHS p+q=(2sinA/1+casA+sinA) + (casA/1+sinA) ={2sinA(1+sinA) + cosA(1+casA+sinA)} / (1+sinA+casA)(1+sinA) =(2sinA+2sinAsinA + casA+casAcasA+casAsinA) / (1+sinA+casA)(1+sinA) ={2sinA+2sinAsinA...
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Given that p = 2sinA /1+cosA+sinA and q = cosA / 1+sinA We have to prove p+q = 1 Proof: LHS p+q=(2sinA/1+casA+sinA) + (casA/1+sinA) ={2sinA(1+sinA) + cosA(1+casA+sinA)} / (1+sinA+casA)(1+sinA) =(2sinA+2sinAsinA + casA+casAcasA+casAsinA) / (1+sinA+casA)(1+sinA) ={2sinA+2sinAsinA + casAcasA +casA(1+sinA)} / (1+sinA+casA)(1+sinA) ={2sinA+sinAsinA+1+casA(1+sinA) } / (1+sinA+casA)(1+sinA) ={sinAsinA +sinA+sinA+1 +casA(1+sinA)} / (1+sinA+casA)(1+sinA) ={sinAsinA+sinA +(sinA+1)(1+casA)} / (1+sinA+casA)(1+sinA) ={sinA(sinA+1) +(sinA+1)(casA+1)} / (1+sinA+casA)(1+sinA) ={(sinA+1)(sinA+casA+1)} / (1+sinA+casA)(1+sinA) =1 RHS Hence Proved. read less
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Tutor

1-q = / (1+sinA) multiplying numerator and denominator by (1+sinA+cosA) = / Simplifying by opening the brackets = / () = 2sinA / = p So, q+p=1 NOTE:- sinA not equal to -1
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

p + q = 2sinA /1+ cos A + sin A + cosA / 1+ sin A = 2sinA + 2 sin2A + cos A + cos2A + cos A.sin A / (1+cos A +sin A) (1 + sinA) = (1+sinA+cosA) + (sin2A + sinA + sinA.cosA) / (1+cosA+sinA) (1+sinA) = (1+sinA+cosA)(1+sinA) / (1+sinA+cosA) (1+sinA) = 1
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p+q=(2sin A)/(1+sinA+cosA) +((cosA/1+sinA))..........TAKE LCM AND MULTIPLY D VALUE IN NUMERATOR AND DENOMINATOR.......D NUMERATOR AND DENOMINATOR WILL COME SAME...AND HENCE 1
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p+q =2sinA/1+cosA +sinA +cosA/1+sinA= STEP-1=2sinA+2sin^2A+cosA+cos^2A+sinAcosA/(1+sinA+cosA)(1+sinA) STEP-2= 2sinA+1+sin^2A+cosA+sinAcosA/(1+sinA+cosA)(1+sinA) {since cos^2A=1-sin^2A} STEP-3= (sinA+cosA+1)+sinA+sin^2A+sinAcosA/(1+sinA+cosA)(1+sinA) {2sinA=sinA+sinA} STEP-4= (1+sinA+cosA)+sinA(1+sinA+cosA)/(1+sinA+cosA)(1+sinA) STEP-5=...
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p+q =2sinA/1+cosA +sinA +cosA/1+sinA= STEP-1=2sinA+2sin^2A+cosA+cos^2A+sinAcosA/(1+sinA+cosA)(1+sinA) STEP-2= 2sinA+1+sin^2A+cosA+sinAcosA/(1+sinA+cosA)(1+sinA) {since cos^2A=1-sin^2A} STEP-3= (sinA+cosA+1)+sinA+sin^2A+sinAcosA/(1+sinA+cosA)(1+sinA) {2sinA=sinA+sinA} STEP-4= (1+sinA+cosA)+sinA(1+sinA+cosA)/(1+sinA+cosA)(1+sinA) STEP-5= (1+sinA+cosA)(1+sinA)/(1+sinA+cosA)(1+sinA)=1 HENCE PROVED NOTE:if any doubt is there in d solution cn text me thnk u read less
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B Tech ( NIT Jamshedpur)

write as SInA= 2sinA/2cosA/2, 1+cosA=2cos2A & CosA=cos2A/2-Sin2A/2
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p+q = (2sinA /1+cosA+sinA) + (CosA/1+sinA) Taking LCM of denominators we get Denominator as (1+cosA+sinA)(1+sinA) Therefore, Numerator becomes (2SinA(1+sinA) + CosA(1+cosA+sinA) =( 2SinA+2Sin^2A+CosA+Cos^2A+CosASinA) = 2SinA+Sin^2A+Sin^2A+Cos^2A+CosA+CosASinA =2SinA+Sin^2A+1+CosA+cosASinA =(SinA+1)(SinA+1)...
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p+q = (2sinA /1+cosA+sinA) + (CosA/1+sinA) Taking LCM of denominators we get Denominator as (1+cosA+sinA)(1+sinA) Therefore, Numerator becomes (2SinA(1+sinA) + CosA(1+cosA+sinA) =( 2SinA+2Sin^2A+CosA+Cos^2A+CosASinA) = 2SinA+Sin^2A+Sin^2A+Cos^2A+CosA+CosASinA =2SinA+Sin^2A+1+CosA+cosASinA =(SinA+1)(SinA+1) +CosA(SinA+1) =(SinA+1)(SinA+CosA+1) Numerator /Denominator = 1=> hence proved Note: Sin^2A => read as Sine square A 2Sin^2A = Sin^2A + Sin^2A Sin^2A + Cos^2A = 1 read less
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The solution given by Shalini is the right way of solving the problem.
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