हिंदी

Cos⁡𝜃⋅cos⁡(90∘−𝜃)/tan⁡(90∘−𝜃) is equivalent to ______.

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प्रश्न

\[\frac{\cos \theta \cdot \cos(90^\circ - \theta)}{\tan(90^\circ - \theta)}\] is equivalent to ______.

विकल्प

  • cos2 θ − 1

  • sin2 θ

  • sin2 θ - cos2 θ

  • sin2 θ − 1

MCQ
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उत्तर

\[\frac{\cos \theta \cdot \cos(90^\circ - \theta)}{\tan(90^\circ - \theta)}\] is equivalent to sin2 θ.

Explanation: 

Solving,

\[\Rightarrow \frac{\cos \theta \cdot \cos(90^\circ - \theta)}{\tan(90^\circ - \theta)}\]

\[\Rightarrow \frac{\cos \theta \cdot \sin \theta}{\cot \theta}\]

\[\Rightarrow \frac{\cos \theta \cdot \sin \theta}{\frac{\cos \theta}{\sin \theta}}\]

\[\Rightarrow \frac{\cos \theta \cdot \sin \theta \cdot \sin \theta}{\cos \theta}\]

\[\Rightarrow \sin^2 \theta.\]

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अध्याय 21: Trigonometrical Identities - EXERCISE 21 (C) [पृष्ठ ३३७]

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सेलिना Concise Mathematics [English] Class 10 ICSE
अध्याय 21 Trigonometrical Identities
EXERCISE 21 (C) | Q 1. (e) | पृष्ठ ३३७
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