CUET Preparation Today
If cosθ + cos2θ = 1, then find the value of \(\sqrt {sin^4θ+cos^2θ}\) |
2 cosθ \(\sqrt {2}\)cosθ \(\sqrt {2}\)tanθ \(\frac{\sqrt {2}}{3}\)secθ |
\(\sqrt {2}\)cosθ |
⇒ cosθ = 1- cos2θ = sin2θ ⇒ cosθ = sin2θ and cos2θ = sin4θ Now, ⇒ \(\sqrt {sin^4θ+cos^2θ}\) = \(\sqrt {cos^2θ+cos^2θ}\) = \(\sqrt {2}\)cosθ |
