If $0 < \theta < \frac{\pi}{2}$ and $\sin \theta \cos \theta=\frac{12}{25}$, then $\sin ^4 \theta+\cos ^4…
If $0 < \theta < \frac{\pi}{2}$ and $\sin \theta \cos \theta=\frac{12}{25}$, then $\sin ^4 \theta+\cos ^4 \theta$ is equal to
- $\frac{327}{625}$
- $\frac{337}{625}$
- $\frac{347}{625}$
- $\frac{340}{625}$
Solution
Given $0 < \theta < \frac{\pi}{2}$ and $\sin \theta \cdot \cos \theta=\frac{12}{25}$
To Find $\sin ^4 \theta+\cos ^4 \theta=$ ?
$
\begin{gathered}
\sin ^4 \theta+\cos ^4 \theta=\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2 \\
\left\{\because a^2+b^2=(a+b)^2-2 a b\right\} \\
=\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2(\sin \theta \cdot \cos \theta)^2 \\
=(1)^2-2\left(\frac{12}{25}\right)^2=1-\frac{288}{625}=\frac{337}{625}
\end{gathered}
$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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