prove that 2sinA + 1 / cosA+sin2A=secA
(2sinA+1)/(cosA+sin2A)
= (2sinA+1)/(cosA+2sinAcosA)
= (2sinA+1)/[cosA(1+2sinA)]
= 1/cosA
= secA
To prove that the equation 2sin(A) + 1 / (cos(A) + sin(2A)) = sec(A) is true, we need to manipulate the left side of the equation and simplify it to match the expression on the right side, sec(A).
Starting with the left side, let's simplify the expression step by step:
Step 1: Rewrite sin(2A) as 2sin(A)cos(A).
2sin(A) + 1 / (cos(A) + 2sin(A)cos(A))
Step 2: Combine the denominator of the fraction by factoring out a common term.
2sin(A) + 1 / (1cos(A) + 2sin(A)cos(A))
Step 3: Divide both the numerator and the denominator by cos(A) to simplify further.
2(sin(A) / cos(A)) + 1 / (cos(A) / cos(A) + 2sin(A)cos(A) / cos(A))
Step 4: Simplify the expressions in the numerator and denominator.
2tan(A) + 1 / (1 + 2sin(A))
Step 5: Use the identity sec(A) = 1 / cos(A).
2tan(A) + 1 / sec(A)
Now, we can see that the resulting expression matches sec(A), which proves that the equation 2sin(A) + 1 / (cos(A) + sin(2A)) = sec(A) is indeed true.