AprblmInTrigo

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

Find $C$

The answer is 1.

1 solution

Soumo Mukherjee
Oct 13, 2014

$\\\left( 1+\tan { \alpha /2 } -\sec { \alpha /2 } \right) \left( 1+\tan { \alpha } +\sec { \alpha /2 } \right) ={ \left( 1+\tan { \alpha /2 } \right) }^{ 2 }-\sec ^{ 2 }{ \alpha /2 } \\ =1+\tan ^{ 2 }{ \alpha /2 } +2\tan { \alpha } /2-\sec ^{ 2 }{ \alpha } /2\\ =\left[ \cos ^{ 2 }{ \alpha /2 } +\tan ^{ 2 }{ \alpha /2 } \cos ^{ 2 }{ \alpha /2 } +2\tan { \alpha /2 } \cos ^{ 2 }{ \alpha /2 } -1 \right] /\left[ \cos ^{ 2 }{ \alpha /2 } \right] \\ =\left[ \left( \cos ^{ 2 }{ \alpha /2 } +\sin ^{ 2 }{ \alpha /2 } \right) +2\sin { \alpha /2 } \cos { \alpha /2 } -1 \right] /\left[ \cos ^{ 2 }{ \alpha /2 } \right] \\= \left[ 2\sin { \alpha /2 } \cos { \alpha /2 } \right] /\left[ \cos ^{ 2 }{ \alpha /2 } \right] =\sin { \alpha \sec ^{ 2 }{ \alpha /2 } } =l=1.l$ Also Visit

AprblmInTrigo

The answer is 1.

1 solution

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