Question

[( tanA / (secA - 1 )] + [( tanA / (secA - 1 )] = 2cosecA ..
Do this by taking LHS or RHS

Answer 1

$\: Given \: \: Question \: \: Is \: \\ \\ \frac{ \tan(x) }{ \sec(x) - 1} + \frac{ \tan(x) }{ \sec(x) +1} \\ \\ \\ Answer \: \: \\ \\ lhs\\ \\ Replace \: \: \: \: \: \tan(x) \: \: by \: \: \: \frac{ \sin(x) }{ \cos(x) } \\ and \\ \sec(x) \: \: \: \: by \: \: \: \: \frac{1}{ \cos(x) } \\ \\ \\ \frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{1}{ \cos(x) } - 1 } + \frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{1}{ \cos(x) } + 1 } \\ \\ \\ \frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{1 - \cos(x) }{ \cos(x) } } + \frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{1 + \cos(x) }{ \cos(x) } } \\ \\ \\ \frac{ \sin(x) }{1 - \cos(x) } + \frac{ \sin(x) }{1 + \cos(x) } \\ \\ \\ \frac{ \sin(x)(1 + \cos(x) ) + \sin(x)(1 - \cos(x) )}{1 {}^{2} - \cos {}^{2} (x) } \\ \\ \\ \frac{ \sin(x) + \sin(x) \cos(x) + \sin(x) - \sin(x) \cos(x) }{ \sin {}^{2} (x) } \\ becoz \: \: \: \: \: 1 - \cos {}^{2} (x) = \sin {}^{2} (x) \\ \\ \\ \frac{2 \sin(x) }{ \sin {}^{2} (x) } \\ \\ \\ \frac{2}{ \sin(x) } \\ \\ \\ 2cosec(x) \: \: \: \: Hence \: \: \: \: proved \: \: \: \\ \: \\ \\ \\ NOTE \: \: \: \: \: \: \: \: here \: \: \: A \: \: \: \: = \: \: \: x \:$

Answer 2

$\dfrac{tanA}{secA \: - \: 1}$ + $\dfrac{tanA}{secA\:+\:1}$ = 2cosecA

_______________[GIVEN]

$\underline{\bold{Method\:1.}}$

• Taking L.H.S.

=> $\dfrac{tanA}{secA \: - \: 1}$ + $\dfrac{tanA}{secA\:+\:1}$

=> $\dfrac{tanA(secA \: + \: 1) \: + \: tanA(seca \: - \: 1)}{(secA \: - \: 1) \: (secA \: + \: 1)}$

=> $\dfrac{tanA(secA \: + \: 1 \: + \: secA\: - \: 1)}{ {sec}^{2}A \: - \: 1 }$

=> $\dfrac{tanA(secA\: + \: 1 \: + \: secA\: - \: 1)}{ {tan}^{2}A}$

=> $\dfrac{tanA(2secA)}{{tan}^{2}A}$

=> $\dfrac{2 \frac{1}{cosA} }{ \frac{sinA}{cosA} }$

=> $\dfrac{2}{sinA}$

=> 2 cosecA

• L.H.S. = R.H.S.

2 cosecA = 2 cosecA

Hence, proved.

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» Used formulas :

• sec²A - 1 = tan²A

=> sec²A - tan²A = 1

• secA = $\dfrac{1}{cosA}$

• tanA = $\dfrac{sinA}{cosA}$

• cosecA = $\dfrac{1}{sinA}$

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$\underline{\bold{Method\:2.}}$

• Taking L.H.S.

=> $\dfrac{tanA}{secA \: - \: 1}$ + $\dfrac{tanA}{secA\:+\:1}$

=> $\dfrac{tanA(secA \: + \: 1) \: + \: tanA(seca \: - \: 1)}{(secA \: - \: 1) \: (secA \: + \: 1)}$

=> $\dfrac{tanA \: secA \: + \: tanA \: + \: tanA \: secA \: - \: tanA}{(secA\: - \: 1)(secA \: + \: 1)}$

=> $\dfrac{2\:tanA \: secA}{ {sec}^{2}A \: - \: 1}$

=> $\dfrac{2\:tanA\:secA}{{tan}^{2}A}$

=> $\dfrac{2\:secA}{tanA}$

=> 2 secA × $\dfrac{1}{tanA}$

=> $\dfrac{2}{cosA}$ × $\dfrac{1}{ \frac{sin}{cos} }$

=> $\dfrac{2}{cosA}$ × $\dfrac{cosA}{sinA}$

=> $\dfrac{2}{sinA}$

=> 2 cosecA

• L.H.S. = R.H.S.

2 cosecA = 2 cosecA

Hence, proved.

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