हिंदी

Evaluate: ( Sin 77 ° Cos 13 ° ) 2 + ( Cos 77 ° Sin 13 ° ) 2 − 2 Cos 2 45 °

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प्रश्न

Evaluate: `((sin 77°)/(cos 13°))^2 + ((cos 77°)/(sin 13°))^2 - 2 cos^2 45°`

योग
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उत्तर

`((sin 77°)/(cos 13°))^2 + ((cos 77°)/(sin 13°))^2 - 2 cos^2 45°`

= `(sin(90° - 13°)/(cos 13°))^2 + (cos(90° - 13°)/(sin 13°))^2 - 2 (cos 45°)^2`

= `((cos 13°)/(cos 13°))^2 + ((sin 13°)/(sin 13°))^2 - 2 (1/sqrt(2))^2`

= `(1)^2 + (1)^2 - 2 xx (1)/(2)`

= 1 + 1 - 1
= 1

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अध्याय 21: Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] - EXERCISE 21 (F) [पृष्ठ ३३४]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 21 Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]
EXERCISE 21 (F) | Q 3. (v) | पृष्ठ ३३४
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