मराठी

Evaluate: Cot 2 41 ° Tan 2 49 ° − 2 Sin 2 75 ° Cos 2 15 ° - Mathematics

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प्रश्न

Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`

बेरीज

उत्तर १

`(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`

= `[cot(90°- 49°)]^2/(tan^2 49°)- 2[sin(90° - 15°)]^2/(cos^2 15°)`

= `(tan^2 49°)/(tan^2 59°) - 2 (cos^2 15°)/(cos^2 15°)`

= 1 - 2
= -1

shaalaa.com

उत्तर २

`(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`

= `[cot(90°- 49°)]^2/(tan^2 49°)- 2[sin(90° - 15°)]^2/(cos^2 15°)`

= `(tan^2 49°)/(tan^2 49°) - 2 (cos^2 15°)/(cos^2 15°)`

= 1 - 2
= -1

shaalaa.com
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पाठ 25: Complementary Angles - Exercise 25 [पृष्ठ ३१०]

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सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 25 Complementary Angles
Exercise 25 | Q 6.7 | पृष्ठ ३१०
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