Question

it's urgent please answer this quetsion ​

Answer 1

$\bf{ \mathcal Prove \: that \: \: \bigg[ 1-\bigg \{1-(1-p²)^{-1}\bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}$

Solution

${ \mathcal{\: \: \mapsto \: \: \bigg[ 1-\bigg \{1-(1-p²)^{-1}\bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}}$

${ \sf\: \: \mapsto \: \: \bigg[ 1-\bigg \{1-( {1}^{2} -p²)^{-1}\bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}$

${\sf \: \: \mapsto \: \: \bigg[ 1-\bigg \{1- \{(1 + p)(1 - p) \}^{-1}\bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[ 1-\bigg \{1- \dfrac{1}{(1 + p)(1 - p)} \bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[1 - \bigg \{ \dfrac{(1 + p)(1 - p) - 1}{(1 + p)(1 - p)} \bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[1 - \bigg \{ \dfrac{ {1}^{2} - {p}^{2} - 1}{(1 + p)(1 - p)} \bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[1 - \bigg \{ \dfrac{ - p \times - p - 1 + 1}{(1 + p)(1 - p)} \bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[1 - \bigg \{ \dfrac{ {p}^{2} }{(1 + p)(1 - p)} \bigg\} ^{-1} \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[1 - \dfrac{ (1 + p)(1 - p) }{ {p}^{2} } \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[1 + \dfrac{ (1 - p)(1 + p) }{ {p}^{2} } \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[ \dfrac{ {p}^{2} + {1}^{2} - {p}^{2} }{ {p}^{2} } \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: \bigg[ \dfrac{ 1 }{ {p}^{2} } \bigg]^{-\dfrac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: {p} ^{2 \times \frac{1}{2} } = p}$

${ \sf\: \: \mapsto \: \: p = p}$

${\bf \mathcal{ \: \: \mapsto \: \: LHS = RHS }_{ \pink{_{_{_{_{ Proved }}}}}}}$