proove point le lo bhai yaaaaararraar
Answers
naku answer idhi vachindhi....so u please understood the problem.
Answer:
Consider LHS,
\rm :\longmapsto\:\dfrac{ {a}^{ - 1} }{ {a}^{ - 1} + {b}^{ - 1} } + \dfrac{ {a}^{ - 1} }{ {a}^{ - 1} - {b}^{ - 1} }:⟼
a
−1
+b
−1
a
−1
+
a
−1
−b
−1
a
−1
\rm \: = \: \: {a}^{ - 1} \bigg( \dfrac{1}{ {a}^{ - 1} + {b}^{ - 1} } + \dfrac{1}{ {a}^{ - 1} - {b}^{ - 1} } \bigg)=a
−1
(
a
−1
+b
−1
1
+
a
−1
−b
−1
1
)
\rm \: = \: \: \dfrac{1}{a} \bigg( \dfrac{1}{ \dfrac{1}{a}+ \dfrac{1}{b} } + \dfrac{1}{ \dfrac{1}{a} - \dfrac{1}{b} } \bigg)=
a
1
(
a
1
+
b
1
1
+
a
1
−
b
1
1
)
\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \because\: {x}^{ - 1} = \dfrac{1}{x} }}
∵x
−1
=
x
1
\rm \: = \: \: \dfrac{1}{a}\bigg(\dfrac{1}{\dfrac{b + a}{ab} } + \dfrac{1}{\dfrac{b - a}{ab} } \bigg)=
a
1
(
ab
b+a
1
+
ab
b−a
1
)
\rm \: = \: \: \dfrac{1}{a}\bigg(\dfrac{ab}{b + a} + \dfrac{ab}{b - a} \bigg)=
a
1
(
b+a
ab
+
b−a
ab
)
\rm \: = \: \: \dfrac{ \cancel a \: b}{ \cancel a}\bigg(\dfrac{1}{b + a} + \dfrac{1}{b - a} \bigg)=
a
a
b
(
b+a
1
+
b−a
1
)
\rm \: = \: \: b\bigg(\dfrac{b - a + b + a}{(b + a)(b - a)} \bigg)=b(
(b+a)(b−a)
b−a+b+a
)
\rm \: = \: \: b\bigg(\dfrac{2b}{(b + a)(b - a)} \bigg)=b(
(b+a)(b−a)
2b
)
\rm \: = \: \: \dfrac{2 {b}^{2} }{ {b}^{2} - {a}^{2} }=
b
2
−a
2
2b
2
Hence, Proved