1-sina = ctga - cosa / ctga ?

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Guest Dec 3, 2014

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Best Answer

I think this is meant to be

$$1-\sin a=\frac{\cot a - \cos a}{\cot a}$$

$$\frac{\cot a - \cos a}{\cot a}=1-\frac{\cos a}{\cot a}=1-\frac{\cos a}{\cos a/\sin a}=1-\sin a$$

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Alan Dec 5, 2014

Alan · Answer 1 · 2014-12-05T01:00:36

I think this is meant to be

$$1-\sin a=\frac{\cot a - \cos a}{\cot a}$$

$$\frac{\cot a - \cos a}{\cot a}=1-\frac{\cos a}{\cot a}=1-\frac{\cos a}{\cos a/\sin a}=1-\sin a$$

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CPhill · Answer 2 · 2014-12-05T11:34:36

1-sina = ctga - cosa / ctga ???

1 - sin a = cos a / sin a - cos a / (cos a / sin a) ???

1 - sin a = cos a / sin a - sina add sin a to both sides

1 = cos a / sin a ????

Nope.......not an identity.......