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Sec 50º Sin 40° + Cos 40º Cosec 50º - Mathematics

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Question

Evaluate the following 

sec 50º sin 40° + cos 40º cosec 50º 

Solution

We have,

sec50º sin40º + cos40º cosec50º

= sec(90º – 40º) sin40º + cos40º cosec(90º – 40º)

= cosec40º sin40º + cos40ºsec40º

`=\frac{\sin 40^\text{o}}{\sin 40^\text{o}}+\frac{\cos 40^\text{o}}{\cos40^\text{o}}`

= 1 + 1 = 2

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APPEARS IN

 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 10: Trigonometric Ratios
Ex. 10.3 | Q: 2.11 | Page no. 53
 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 10: Trigonometric Ratios
Ex. 10.3 | Q: 2.11 | Page no. 53
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Solution Sec 50º Sin 40° + Cos 40º Cosec 50º Concept: Trigonometric Ratios of Some Specific Angles.
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