If cos2 a - sinº a = tan’B, show that cos- sinB = tan a

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( cos ^{2} alpha-sin ^{2} alpha=tan ^{2} beta cdot pi ) ( frac{1 beta}{b text { th side }} ) ( begin{aligned} cos ^{2} alpha &-sin ^{2} alpha-1=frac{sin ^{2} beta}{cos ^{2} beta}-1 &-2 sin ^{2} alpha=frac{sin ^{2} beta-cos ^{2} beta}{cos ^{2} beta} &=frac{2 sin ^{2} alpha}{sec ^{2} beta}=sin ^{2} beta-cos ^{2} beta end{aligned} ) ( frac{sin ^{2} alpha}{cos ^{2} alpha+sin ^{2} alpha}=sin ^{2} beta-cos ^{2} betaleft[frac{-2 sin ^{2} alpha}{1+tan ^{2} beta}=sin ^{2} beta-cos ^{2} betaright. ) ( =cos ^{2} beta cdot sin ^{2} beta mid< ) 7 ( tan ^{2} alpha=cos ^{2} beta-sin ^{2} beta )

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