AprblmInTrigo

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If

(sinαsec2α2=l)\biggl(\sin\alpha \sec ^{ 2 }{ \frac { \alpha }{ 2 } } =l\biggr) and (1+tanα2secα2)(1+tanα2+secα2)=Cl\left( 1+\tan { \frac { \alpha }{ 2 } } -\sec { \frac { \alpha }{ 2 } } \right) \left( 1+\tan { \frac { \alpha }{ 2 } } +\sec { \frac { \alpha }{ 2 } } \right) =Cl

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