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Evaluate:

\(\frac{sin\, 21°}{cos\, 69°}\)

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Best answer

We have, 

\(\frac{sin 21^o}{cos 69^o }\)

= \(\frac{sin (90^o – 69^o)}{cos 69^o }\)

= \(\frac{cos 69^o}{ cos69^o }\)

= 1 [∵ sin (90 – θ) = cos θ]

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