Math, asked by aayushi27, 1 year ago

plz solve both of them

Attachments:

Answers

Answered by jaya1012
3
According to given sum,

(i)

lhs = \frac{ {a}^{ - 1} }{ {a}^{ - 1} + {b}^{ - 1} } + \frac{ {a}^{ - 1} }{ {a}^{ - 1} - {b}^{ - 1} }

 = > \: \frac{1}{a} ( \frac{1}{ {a}^{ - 1} + {b}^{ - 1} } ) + \frac{1}{a}( \frac{1}{ {a}^{ - 1} - {b}^{ - 1} } )

 = > \: \frac{1}{1 + \frac{a}{b} } + \frac{1}{1 - \frac{a}{b} }

 = > \: \frac{b}{a + b} + \frac{b}{b - a}

 = > \: b( \frac{1}{a + b} + \frac{1}{b - a} )

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

 = > \: b( \frac{2b}{ {b}^{2} - {a}^{2} } )

 = > \: \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} }

LHS = RHS .

Hence, proved.

(ii)

lhs = \frac{ {a} + {b} + {c} }{ {a}^{ - 1}. {b}^{ - 1} + {b}^{ - 1}. {c}^{ - 1} + {a}^{ - 1} . {c}^{ - 1} }

 = > \: \frac{a + b + c}{ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} }

Taking the LCM of the denominators.

 = > \: \frac{a + b + c}{ \frac{a + b + c}{abc} }

 = > \: abc( \frac{a + b + c}{a + b + c} )

=> abc.

LHS = RHS .

Hence, proved

:-)Hope it helps u .
Similar questions