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 cosθ and sinθ . Since we now know our direction, lets sail mate.
(1+tan2α−sec2α)(1+tanα+sec2α)=(1+tan2α)2−sec22α=1+tan22α+2tanα/2−sec2α/2=[cos22α+tan22αcos22α+2tan2αcos22α−1]/[cos22α]=[(cos22α+sin22α)+2sin2αcos2α−1]/[cos22α]=[2sin2αcos2α]/[cos22α]=sinαsec22α=l
Hence the answer is 1
#Trigonometry
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Please don't use so weird formatting. Only give the \ ( .... \ ) wrapping around the Math Terms and not around the words.
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Thanx buddy! Nxt time I will.
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I've edited this, take a note of the changes and use what you understand in future.
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