5cosθ+12sinθ=13, tanθ=?

প্রশ্ন

5\cos\theta + 12\sin\theta = 13, \tan\theta=?

উত্তর

\tan\theta = \dfrac{12}{5}

ব্যাখ্যা / সমাধান

5\cos\theta + 12\sin\theta = 13

\Rightarrow (5\cos\theta + 12\sin\theta)^2 = (13)^2 [উভয়পক্ষে বর্গ করে পাই]

\Rightarrow 25\cos^2\theta + 120\sin\theta\cos\theta + 144\sin^2\theta = 169

\Rightarrow 25\cos^2\theta + 120\sin\theta\cos\theta + 144\sin^2\theta = 169\times 1 = 169(\sin^2\theta + \cos^2\theta) [\because \sin^2\theta + \cos^2\theta = 1]

\Rightarrow 25\cos^2\theta + 120\sin\theta\cos\theta + 144\sin^2\theta = 169\sin^2\theta + 169\cos^2\theta

\Rightarrow 120\sin\theta\cos\theta = 25\sin^2\theta + 144\cos^2\theta

\Rightarrow 25\sin^2\theta - 120\sin\theta\cos\theta + 144\cos^2\theta = 0

\Rightarrow (5\sin\theta)^2 - 2\cdot5\sin\theta\cdot 12\cos\theta + (12\cos\theta)^2 = 0

\Rightarrow (5\sin\theta - 12\cos\theta)^2 = 0

\Rightarrow 5\sin\theta - 12\cos\theta = 0

\Rightarrow 5\sin\theta = 12\cos\theta

\Rightarrow \dfrac{\sin\theta}{\cos\theta} = \dfrac{12}{5}

\Rightarrow \tan\theta = \dfrac{12}{5}

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