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Solution 4.3:8a

From Förberedande kurs i matematik 1

Revision as of 08:16, 10 October 2008 by Tek (Talk | contribs)
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We rewrite tanv on the left-hand side as sinvcosv, so that

tan2v=sin2vcos2v.

If we then use the Pythagorean identity

cos2v+sin2v=1

and rewrite cos2v in the denominator as 1−sin2v, we get what we are looking for on the right-hand side. The whole calculation is

tan2v=sin2vcos2v=sin2v1−sin2v.
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