Practicing Success
If θ is an acute angle and sin θ = cos θ, then the value of $2 tan^2 θ + sin^2 θ - 1 $ is equal to : |
-7 1 3 $\frac{3}{2}$ |
$\frac{3}{2}$ |
sinθ = cosθ { we know, sinA = cosB Iff A + B = 90º } So, θ + θ= 90º θ= 45º Now, 2tan²θ + sin²θ - 1 = 2tan²45º + sin²45º - 1 = 2 + ( \(\frac{1}{√2}\) )² - 1 = 2 + \(\frac{1}{2}\) - 1 = \(\frac{3}{2}\) |