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Evaluate the Following : (Sin 72° + Cos 18°) (Sin 72° − Cos 18°) - CBSE Class 10 - Mathematics

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Question

Evaluate the following :

(sin 72° + cos 18°) (sin 72° − cos 18°)

Solution

We have to find: (sin 72° + cos 18°) (sin 72° − cos 18°)

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

sin 72° + cos 18°) (sin 72° − cos 18°) = `(sin 72^@)^2 - (cos 18^@)^2`

`= [sin (90^@ - 18^@)]^2 - (cos 18^@)^2`

`= (cos 18^@)^2 - (cos 18^@)^2`

`= cos^2 18^@ - cos^2 18^@`

= 0

So value of `(sin 72^@ + cos 18^@)(sin 72^@ - cos 18^@)` is 0

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Solution Evaluate the Following : (Sin 72° + Cos 18°) (Sin 72° − Cos 18°) Concept: Trigonometric Ratios of Some Specific Angles.
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