Practicing Success
Practicing Success
CUET Preparation Today
$(sin θ + cos θ)^2 = 2, 0° < θ < 90°$, then the value of θ is : |
0 $\frac{\pi}{2}$ $\pi$ $\frac{\pi}{4}$ |
$\frac{\pi}{4}$ |
( sinθ + cosθ )² = 2 sin²θ + cos²θ + 2sinθ.cosθ = 2 { we know, sin²θ + cos²θ = 1 } 1 + 2sinθ.cosθ = 2 { 2sinθ.cosθ = sin2θ } sin2θ = 1 sin2θ = sin90º { sin90º = 1 } 2θ = 90º θ = 45º θ = \(\frac{π }{4}\) |
