Question

Simplify :

$\frac{ \sqrt{1 + a} }{ \sqrt{1 + a} - \sqrt{1 - a} } - \: \frac{1 - a}{ \sqrt{1 - {a}^{2} } \: + \: a - 1 }$

a is not equal to 0.​

Answer 1

Answer:

$\boxed{\huge{1}}$

Step-by-step explanation:

$\bold{To \: find}\longrightarrow \: \: \: value \: of \\ \dfrac{ \sqrt{1 + a} }{ \sqrt{1 + a} - \sqrt{1 - a} } - \dfrac{(1 - a)}{ \sqrt{1 - a^{2} } + (a - 1)}$

$\bold{Concept \: used}\longrightarrow \\ 1)( \sqrt{x} \: ) ^{2} = x$

$\bold{Soution}\longrightarrow \\ \dfrac{ \sqrt{1 + a} }{ \sqrt{1 + a} - \sqrt{1 - a} } - \dfrac{1 - a}{ \sqrt{1 - {a}^{2} } + (a - 1)}$

$= \dfrac{ \sqrt{1 + a} }{ \sqrt{1 + a} - \sqrt{1 - a} } - \dfrac{( \: \sqrt{1 - a} \: ) ^{2} }{ \sqrt{(1) ^{2} - ({a})^{2} } - (1 - a) }$

$= \dfrac{ \sqrt{1 + a} }{ \sqrt{1 + a} - \sqrt{1 - a} } - \dfrac{ ( \: { \sqrt{1 - a} \: ) }^{2} }{ \sqrt{(1 + a) \: (1 - a)} - {( \: \sqrt{1 - a} \: ) }^{2} }$

$= \dfrac{ \sqrt{1 + a} }{ \sqrt{1 + a} - \sqrt{1 - a} } - \dfrac{ ( \: { \sqrt{1 - a} \: ) }^{2} }{ \sqrt{(1 + a) \: (1 - a)} - {( \: \sqrt{1 - a} \: ) }^{2} }$

$taking \: \sqrt{1 - a} \: common \: in \\ second \: term \: numerator \\ = \dfrac{ \sqrt{1 + a} }{ \sqrt{1 + a} - \sqrt{1 - a} } - \dfrac{ { (\sqrt{1 - a} })^{2} }{ \sqrt{1 - a} \: ( \sqrt{1 + a} - \sqrt{1 - a} \: )}$

$\sqrt{1 - a} \: cancel \: out \: from \: second \: term \: numerator \: and \: denominator$

$= \dfrac{ \sqrt{1 + a} }{ \sqrt{1 + a} - \sqrt{1 - a} } - \dfrac{ \sqrt{1 - a} }{ \sqrt{1 + a} - \sqrt{1 - a} }$

$= \dfrac{ \sqrt{1 + a} - \sqrt{1 - a} }{ \sqrt{1 + a} - \sqrt{1 - a} }$

$= 1$

Answer 2

Answer:

$\mapsto\tt{\dfrac{\sqrt{1 + a}}{ \sqrt{1 + a} - \sqrt{1 - a}} - \dfrac{1 - a}{ \sqrt{1 - {a}^{2}} + a - 1}}$

Solving it,

$\mapsto\tt{\dfrac{\sqrt{1 + a}}{ \sqrt{1 + a} - \sqrt{1 - a}} - \dfrac{ {( \sqrt{1 - a} )}^{2} }{ \sqrt{ {1}^{2} - {a}^{2} } - (1 - a)}}$

By √(x)² = x.

$\mapsto\tt{\dfrac{\sqrt{1 + a}}{ \sqrt{1 + a} - \sqrt{1 - a}} - \dfrac{ {( \sqrt{1 - a} )}^{2} }{ \sqrt{ (1+a)(1-a) } - {(\sqrt{(1 - a)})}^{2}}}$

$\mapsto\tt{\dfrac{\sqrt{1 + a}}{ \sqrt{1 + a} - \sqrt{1 - a}} - \dfrac{\sqrt{1 - a}\times \sqrt{1-a}}{ \sqrt{1-a}(\sqrt{1+a}-\sqrt{1-a})}}$

$\mapsto\tt{\dfrac{\sqrt{1 + a}}{ \sqrt{1 + a} - \sqrt{1 - a}} - \dfrac{ \sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}}$

By, LCM.

$\mapsto\tt\cancel{\dfrac{\sqrt{1+a}-\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}}$

$\mapsto\tt{1}$

$\rule{200}$