1.

Plz solve both of them

Answer»

According to given sum,

(i)

<klux>LHS</klux> = \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 .


Discussion

No Comment Found

Related InterviewSolutions