Math, asked by sukani0s9apatri, 1 year ago

Cos a+sin a=√2sin(90-a) show that sina-cosa=√2cosa

Answers

Answered by ARoy
4
cosa+sina=√2sin(90°-a)
or, cosa+sina=√2cosa [∵, sin(90°-a)=cosa]
or, sina=√2cosa-cosa
or, sina=cosa(√2-1)
or, sina=cosa(√2-1)(√2+1)/(√2+1)
or, √2sina+sina=cosa{(√2)²-(1)²}
or, √2sina+sina=cosa(2-1)
or, sina-cosa=-√2sina
or, cosa-sina=√2sina
Please check the question!
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