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Trigonometric Identities

Prove the fol...

Question

Prove the following trigonometric identities:

$\frac{t a n A}{1 + s e c A} - \frac{t a n A}{1 - s e c A} = 2 c o s e c A$

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Solution

$L H S space equals space fraction numerator tan A over denominator 1 plus s e c A end fraction minus fraction numerator tan A over denominator 1 minus s e c A end fraction equals space fraction numerator up diagonal strike tan A end strike minus tan A s e c A minus up diagonal strike tan A end strike minus tan A s e c A over denominator open parentheses 1 minus s e c A close parentheses open parentheses 1 plus s e c A close parentheses end fraction equals fraction numerator negative 2 tan A s e c A over denominator open parentheses 1 minus s e c squared A close parentheses end fraction equals fraction numerator 2 tan A s e c A over denominator open parentheses s e c squared A minus 1 close parentheses end fraction sin c e space s e c squared A minus tan squared A equals 1 h e n c e equals fraction numerator 2 up diagonal strike tan A end strike s e c A over denominator tan to the power of up diagonal strike 2 end exponent A end fraction equals fraction numerator 2 s e c A over denominator begin display style fraction numerator s e c A over denominator cos e c A end fraction end style end fraction space equals space 2 cos e c A$
since LHS = RHS hence proved

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