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3.Trigonometrical Ratios, Functions and Identities
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यदि $\alpha $ समीकरण $25{\cos ^2}\theta + 5\cos \theta - 12 = 0$, $\pi /2 < \alpha < \pi $ का एक मूल हो, तो $\sin 2\alpha $ का मान होगा
A
$24/25$
B
$ - 24/25$
C
$13/18$
D
$ - 13/18$
Solution
चूँकि $\alpha $ समीकरण $25{\cos ^2}\theta + 5\cos \theta – 12 = 0$ का मूल है।
$\therefore$ $25{\cos ^2}\alpha + 5\cos \alpha – 12 = 0$
$\cos \alpha = \frac{{ – 5 \pm \sqrt {25 + 1200} }}{{50}}$ $ = \frac{{ – 5 \pm 35}}{{50}}$
$\cos \alpha = – 4/5$ $[ \because \pi /2 < \alpha < \pi \Rightarrow \cos \alpha < 0]$
$\therefore $$\sin 2\alpha = 2\sin \alpha \cos \alpha = – 24/25$.
Standard 11
Mathematics
