Question

Prove That

tan^2 A / (secA - 1)^2 = 1 + cosA / 1 - cosA ​

Answer 1

$\huge \boxed{ \bf{Answer}}$


$\bf{To \: Prove \: : \frac{ {tan}^{2} A }{( {sec \: A- 1})^{2} } } = \frac{1 + cosA }{1 - cosA }$


$\bf \underline{ Proof} :$

L.H.S,

$= \: \frac{ {tan}^{2} A }{( {sec \: A- 1})^{2} } \:$

$= \frac{ {sec} \: ^{2} A - 1 }{ {(sec A - 1)}^{2} }$

$= \frac{ {sec} \: ^{2} A - \: {1}^{2} }{ {(sec A - 1)}^{2} } \:\: \: \: \: \: -->(numerator\: in \: the \: form \: {a}^{2} - {b}^{2} )$

$= \frac{(sec A + \: 1)(sec A - \: 1)}{ ({sec A - 1)}(sec A - \: 1)}$

Cancelling (sec A - 1) we get,

$= \frac{(sec A + 1)}{(sec A - 1)}$

$= \frac{ \frac{1}{cos A } + 1 }{ \frac{1}{cos A} - 1}$

$= \frac{ \frac{1 + cos A}{cos A} }{ \frac{1 - cos A }{cosA } }$

Cancelling (cos A) we get,

$= \frac{1 + cosA }{1 - cos A }$

$= R.H.S$


Hence proved.



$\bf \underline{Above \: used \: formulas : }$

$i > > \: {tan}^{2} A \: = {sec}^{2} A - 1$

$ii > > \: {a}^{2} - {b}^{2} = (a + b)(a - b)$

$iii > > \: secA = \frac{1}{cosA }$

Answer 2

Step-by-step explanation:

The proof is is in the above....

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