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{{ A-1}{ A+1}}+{{ A+1}{ A-1}}=2 A

Question

$\frac{sec A - 1}{sec A + 1} + \frac{sec A + 1}{sec A - 1} = 2 csc A$

A

True

B

False

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\frac{sec A - 1}{sec A + 1} + \frac{sec A + 1}{sec A - 1} = 2 csc A

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Text solutionVerified

Given

\frac{sec A - 1}{sec A + 1} + \frac{sec A + 1}{sec A - 1}

Mulitiplying and Diving first term by

sec A - 1

Mulitiplying and Diving second term by

sec A + 1

\frac{( sec A - 1 ) ^{2}}{( sec A + 1 ) ( sec A - 1 )} + \frac{( sec A + 1 ) ^{2}}{( sec A - 1 ) ( sec A + 1 )}

\frac{( sec A - 1 ) ^{2}}{( sec ^{2} A - 1 )} + \frac{( sec A + 1 ) ^{2}}{( sec ^{2} A - 1 )}

((a + b) (a - b) = a^{2} - b^{2})

(\frac{sec A - 1}{tan A})^{2} + (\frac{sec A + 1}{tan A})^{2}

(sec^{2} A - 1 = tan^{2} A)

\frac{sec A - 1}{tan A} + \frac{sec A + 1}{tan A}

\frac{2 sec A}{tan A}

\frac{2 ( \frac{1}{cos A} )}{( \frac{sin A}{cos A} )}

\frac{2}{sin A}

2 csc A

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Question Text	$\frac{sec A - 1}{sec A + 1} + \frac{sec A + 1}{sec A - 1} = 2 csc A$
Answer Type	Text solution:1
Upvotes	150