3 tanθ = \(2\sqrt 3 \) sinθ, 0° < θ < 90° అయితే, \(\rm \frac{{\cos e{c^2}2\,\theta + {{\cot }^2}2\,\theta }}{{{{\sin }^2}\,\theta + {{\tan }^2}2\,\theta }}\) విలువ:

Question

Question

This question was previously asked in

SSC CGL Tier 2 Quant Previous Paper 1 (Held On: 29 Jan 2022)

Answer 1

ఇచ్చిన సమస్య:

3 tanθ = \(2√ 3 \) sinθ

ఉపయోగించిన సూత్రం:

tanθ = \(\frac{sinθ }{{cosθ }}\)

cos 30° = \(\frac{√3}{{2}}\), cosec 60° = \(\frac{2}{{√3}}\)

Cot 60° = \(\frac{1}{{√ 3}}\), sin 30° = \(\frac{1}{{2}}\), tan 60° = √3

సాధన:

3 tanθ = \(2√ 3 \) sinθ

⇒ 3 \(\frac{sinθ }{{cosθ }}\) = 2√3 sinθ

⇒ \(\frac{3}{{cosθ }}\) = 2√3

⇒ cosθ = \(\frac{√3}{{2}}\) = cos 30°

⇒ θ = 30°

ఇప్పుడు,

\(\rm \frac{{\cos e{c^2}2\,θ + {{\cot }^2}2\,θ }}{{{{\sin }^2}\,θ + {{\tan }^2}2\,θ }}\) విలువ

⇒ \(\frac{cosec^2 60° + cot^2 60° }{{sin^2 30° + tan^2 60° }}\)

⇒ \(\frac{(\frac{2}{{√3}})^2 \ + \ (\frac{1}{{√3}})^2} {(\frac{1}{{2}})^2 \ + \ (\sqrt3)^2}\)

⇒ \(\frac{\frac{4}{3} + \frac{1}{3}}{\frac{1}{4} + \frac{3}{1}}\) = \(\frac{\frac{5}{3} }{\frac{13}{4} }\)

⇒ \(\frac{{20}}{{39}}\)

∴ అవసరమైన విలువ \(\frac{{20}}{{39}}\).

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