Math, asked by aayushimalvi, 5 months ago

sin 65°/cot25°+cos32°/sin58°-sin28°•sec62°+cosec²30°

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Answered by Officialsakshi

0

Answer:

, It is given that (sin 65/cos25) + ( cos32/sin58 )- (sin28*sec62) + (cosec30* cosec30) Here (sin 65/cos25) = (sin(90–25)/ cos25) .

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