Solution 4.3:8a

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We rewrite \displaystyle \text{tan }v on the left-hand side as \displaystyle \frac{\sin v}{\cos v}, so that


\displaystyle \tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}


If we then use the Pythagorean identity


\displaystyle \cos ^{2}v+\sin ^{2}v=1


and rewrite \displaystyle \text{cos}^{\text{2}}v in the denominator as \displaystyle \text{1}-\text{sin}^{\text{2}}v\text{ }, we get what we are looking for on the right-hand side. The whole calculation is


\displaystyle \tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}=\frac{\sin ^{2}v}{1-\sin ^{2}v}