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Evaluate (sin^2 63°+sin^2 27°)/(cos^2 17°+cos^2 73°) - CBSE Class 10 - Mathematics

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Question

 

Evaluate

`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`

 

Solution

 

`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^@73^@)`

`(= [sin(90^@ - 27^@)]^2+sin^2 27^@)/([cos(90^@ - 73^@)]^2 + cos^2 73^@)`

`= ([cos27^@]^2 + sin^2 27^@)/([sin 73^@]^2 + cos^2 73^@)`

`= (cos^2 27^@ + sin^2 27^@)/(sin^2 73^@+ cos^2 73^@)`

= 1/1 (As sin2A + cos2A = 1)

= 1

 
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APPEARS IN

 NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 8: Introduction to Trigonometry
Ex. 8.40 | Q: 3.1 | Page no. 193
Solution Evaluate (sin^2 63°+sin^2 27°)/(cos^2 17°+cos^2 73°) Concept: Trigonometric Identities.
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