আমাদের টেলিগ্রাম চ্যানেলে যুক্ত হোন

$\frac{\sin \alpha sin11\alpha + sin3\alpha sin7\alpha }{\sin \alpha cos11\alpha + sin3\alpha cos7\alpha }$= $tan8\alpha$

by
$\frac{\sin \alpha sin11\alpha + sin3\alpha sin7\alpha }{\sin \alpha cos11\alpha + sin3\alpha cos7\alpha }$= $tan8\alpha$
Solution: $\frac{\sin \alpha sin11\alpha + sin3\alpha sin7\alpha }{\sin \alpha cos11\alpha + sin3\alpha cos7\alpha }$
= $\frac{2\sin \alpha sin11\alpha + 2sin3\alpha sin7\alpha }{2\sin \alpha cos11\alpha + 2sin3\alpha cos7\alpha }$
= $\frac{\cos (\alpha – 11\alpha ) – cos(\alpha + 11\alpha ) + cos(3\alpha – 7\alpha ) – cos(3\alpha + 7\alpha )}{\sin (\alpha + 11\alpha ) + sin(\alpha – 11\alpha ) + sin(3\alpha + 7\alpha ) + sin(3\alpha – 7\alpha )}$
= $\frac{\cos 10\alpha – cos12\alpha + cos4\alpha – cos10\alpha }{\sin 12\alpha – sin10\alpha + sin10\alpha – sin4\alpha }$ [$\because sin\left( – \theta \right) = – sin\theta$]
= $\frac{\cos 4\alpha – cos12\alpha }{\sin 12\alpha – sin4\alpha }$
= $\frac{2sin\frac{4\alpha + 12\alpha }{2}\sin \frac{12\alpha – 4\alpha }{2}}{2cos\frac{12\alpha + 4\alpha }{2}\sin \frac{12\alpha – 4\alpha }{2}}$
= $\frac{2sin8\alpha sin4\alpha }{2cos8\alpha sin4\alpha }$
= $tan8\alpha $ [Proved]

আরও দেখুন

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