যদি cotθ=2 হয়,তবে 10 sin2θ-6 tan2θ এর মান কোনটি?

প্রদান: \(\cot \theta = 2\)

আমরা জানি, \(\cot \theta = \frac{\cos \theta}{\sin \theta} = 2\)

অর্থাৎ, \(\cos \theta = 2 \sin \theta\)

এখন, \(\sin^2 \theta + \cos^2 \theta = 1\)

অর্থাৎ, \(\sin^2 \theta + (2 \sin \theta)^2 = 1\)

\(\sin^2 \theta + 4 \sin^2 \theta = 1\)

\(5 \sin^2 \theta = 1\)

\(\sin^2 \theta = \frac{1}{5}\)

এখন, \(\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{5} = \frac{4}{5}\)

তাহলে, \(\sin \theta = \pm \frac{1}{\sqrt{5}}\), \(\cos \theta = \pm \frac{2}{\sqrt{5}}\)

অর্থাৎ, \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\pm \frac{1}{\sqrt{5}}}{\pm \frac{2}{\sqrt{5}}} = \frac{1}{2}\)

এখন, \(\sin^2 \theta = \frac{1}{5}\), \(\tan^2 \theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\)

প্রশ্নে দেওয়া: \(10 \sin^2 \theta - 6 \tan^2 \theta\)

অর্থাৎ: \(10 \times \frac{1}{5} - 6 \times \frac{1}{4}\)

\(= 2 - \frac{6}{4}\)

\(= 2 - \frac{3}{2}\)

\(= \frac{4}{2} - \frac{3}{2} = \frac{1}{2}\)

তাই, উত্তরটি \(\frac{1}{2}\)।