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Answers
Prove that
Solution :-
Consider LHS,
Hence, Proved
Additional Information : -
More Identities to know:
- (a + b)² = a² + 2ab + b²
- (a - b)² = a² - 2ab + b²
- a² - b² = (a + b)(a - b)
- (a + b)² = (a - b)² + 4ab
- (a - b)² = (a + b)² - 4ab
- (a + b)² + (a - b)² = 2(a² + b²)
- (a + b)³ = a³ + b³ + 3ab(a + b)
- (a - b)³ = a³ - b³ - 3ab(a - b)
Answer:
\large\underline{\bold{Given \:Question - }}
GivenQuestion−
Prove that
\rm :\longmapsto\:\dfrac{ {a}^{ - 1} }{ {a}^{ - 1} + {b}^{ - 1} } + \dfrac{ {a}^{ - 1} }{ {a}^{ - 1} - {b}^{ - 1} } = \dfrac{2 {b}^{2} }{ {b}^{2} - {a}^{2} } :⟼
a
−1
+b
−1
a
−1
+
a
−1
−b
−1
a
−1
=
b
2
−a
2
2b
2
Solution :-
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
Additional Information : -
\boxed{ \sf{ \: {x}^{m} \times {x}^{n} = {x}^{m + n}}}
x
m
×x
n
=x
m+n
\boxed{ \sf{ \: {x}^{m} \div {x}^{n} = {x}^{m - n}}}
x
m
÷x
n
=x
m−n
\boxed{ \sf{ \: {( {x}^{m} )}^{n} = {x}^{mn} }}
(x
m
)
n
=x
mn
\boxed{ \sf{ \: {x}^{0} = 1}}
x
0
=1
\boxed{ \sf{ \: {x}^{ - y} = \dfrac{1}{ {x}^{y} } }}
x
−y
=
x
y
1
More Identities to know:
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
a² - b² = (a + b)(a - b)
(a + b)² = (a - b)² + 4ab
(a - b)² = (a + b)² - 4ab
(a + b)² + (a - b)² = 2(a² + b²)
(a + b)³ = a³ + b³ + 3ab(a + b)
(a - b)³ = a³ - b³ - 3ab(a - b)