Prove that sin θ + tan θ + cos θ = sec θ
Complete solution pls
with solution = 5 points
For θ = π/4, LHS = √2 + 1 ≠ √2 = RHS, so the equation is false.
Did you mean
sin² (θ) + tan² (θ) + cos² (θ) = sec² (θ)?
This is true because
sin² (θ) + cos² (θ) = 1 = sec² (θ) - tan² (θ).
I don't believe it's true.
sin(pi/4) + tan(pi/4) + cos(pi/4) = [1/sqrt(2)] + 1 + [1/sqrt(2)] = 1 + 2/sqrt(2) = 1 + sqrt(2)
sec(pi/4) = sqrt(2)
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Verified answer
For θ = π/4, LHS = √2 + 1 ≠ √2 = RHS, so the equation is false.
Did you mean
sin² (θ) + tan² (θ) + cos² (θ) = sec² (θ)?
This is true because
sin² (θ) + cos² (θ) = 1 = sec² (θ) - tan² (θ).
I don't believe it's true.
sin(pi/4) + tan(pi/4) + cos(pi/4) = [1/sqrt(2)] + 1 + [1/sqrt(2)] = 1 + 2/sqrt(2) = 1 + sqrt(2)
sec(pi/4) = sqrt(2)