AprblmInTrigo Soln:

The Problem: AprblmInTrigo

\( \sin\alpha \sec ^{ 2 }{ \frac{\alpha }{2} } =l ...(A)\) and \( \left( 1+\tan { \frac{\alpha }{2} } -\sec { \frac{\alpha }{2} } \right) \left( 1+\tan { \frac{\alpha }{2} } +\sec { \frac{\alpha }{2} } \right) =Cl ...(B)\)

The question simply asks for expressing (B) in terms of (A).\ i.e in terms like \(\cos { \theta }\) and \( \sin { \theta }\) . Since we now know our direction, lets sail mate.

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

Hence the answer is \(1\)

Note by Soumo Mukherjee
3 years, 6 months ago

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Please don't use so weird formatting. Only give the \ ( .... \ ) wrapping around the Math Terms and not around the words.

Aditya Raut - 3 years, 6 months ago

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Thanx buddy! Nxt time I will.

Soumo Mukherjee - 3 years, 6 months ago

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I've edited this, take a note of the changes and use what you understand in future.

Aditya Raut - 3 years, 6 months ago

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@Aditya Raut yep.thanx again.

Soumo Mukherjee - 3 years, 6 months ago

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