Question

prove that:
(sin2A+sinA)/(1+cosA+cos2A)=tanA

Answer 1

Answer:-

LHS = (Sin(2A)+Sin(A))/(1+ Cos(2A)+Cos(A))

=(2Sin(A)Cos(A) +Sin(A))/(1+Cos(A)+Cos^2(A)-Sin^2(A))

= Sin(A)(2Cos(A)+1)/(Cos*2(A) + Sin^2(A) + Cos^2(A) - Sin^2(A) + Cos(A))

=Sin(A)(2Cos(A)+1)/(2Cos^2(A) +Cos(A)) = Sin(A)(2Cos(A) + 1)/(Cos(A)(2Cos(A)+1))

=Sin(A)/Cos(A) = Tan(A) = RHS.

Answer 2

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$\dashrightarrow$ LHS= $\LARGE\frac{sin2A+sinA}{1+cosA+cos2A}$

$\dashrightarrow$ $\LARGE\frac{2sinAcosA+sinA}{1+cosA+2cos²A−1}$

$\dashrightarrow$ $\LARGE\frac{2sinAcosA+sinA}{2cos²A+cosA}$

$\dashrightarrow$ $\LARGE\frac{sinA(2cosA+1)}{cosA(2cosA+1)}$

$\dashrightarrow$ $\LARGE\frac{sinA}{cosA}$ = tanA = RHS

$\implies{\rm }$Explanation to the above answer.

Step 1: Copying the L.H.S. from the question.

Step 2: We further expand the multiple angles of sin and cos using the following identities. (sin2A = 2sinAcosA) and (cos2A = 2cos²A -1).

Step 3: On the denominator, as we expand cos2A, we get +1 and -1, which when added results 0. Now, we write the obtained expression.

Step 4: To be able to divide the numerator and denominator in algebra, we need to find the factors. So, we take the common factor in numerator as well as denominator and re-write the expression.

Step 5: After we divide the common factors in the numerator and denominator, they result 1. When multiplying the remaining factors by 1, we get sinA/cosA.

Step 6: (sinA/cosA = tanA)

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