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Evaluate cos 48° − sin 42° - CBSE Class 10 - Mathematics

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Question

Evaluate cos 48° − sin 42°

Solution

We have to find cos 48° − sin 42°

Since `cos (90^@ - theta) = sin theta` So

`cos 48^@ - sin 42^@  = cos (90^@ - 42^@) - sin 42^@`

`= sin 42^@ - sin 42^@`

= 0

So value of `cos 48^@ - sin 42^@` is 0

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 8: Introduction to Trigonometry
Ex.8.30 | Q: 1.3 | Page no. 189
Solution Evaluate cos 48° − sin 42° Concept: Trigonometric Ratios of Some Specific Angles.
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