Question

if cot theta(1+sin theta)=4m and cot theta(1-sin theta)=4n,then prove that (m^2-n^2)^2=mn.
plz frnds help me...plzzzzzzz

plzzzzz....frnds plz...

Answer 1

GIVEN

$= > cot \alpha (1 + \sin \alpha ) = 4m \\ \\ = > cot \alpha (1 - sin \alpha ) = 4n$
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TO PROVE :-

$= > {( {m}^{2} - {n}^{2}) }^{2} = mn$
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FIRST WE SIMPLFY THE RHS

$= > mn = \frac{cot \alpha (1 + sin \alpha )}{4} . \frac{cot \alpha (1 - sin \alpha )}{4} \\ \\ = > mn = \frac{ {cot}^{2} \alpha (1 - {sin}^{2} \alpha ) }{16} \\ \\ = > mn = \frac{ {cot}^{2} \alpha . {cos}^{2} \alpha }{16} \\ \\ = > mn = \frac{ {cos}^{2} \alpha . {cos}^{2} \alpha }{ {sin}^{2} \alpha .16 } \\ \\ = > mn = \frac{ {cos}^{4} \alpha }{16 {sin}^{2} \alpha } \: \: \: \: \: ...[Eq _{1}]$
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NOW WE TAKE LHS :-

$= > {( {m}^{2} - {n}^{2}) }^{2} = {( \frac{ {cot}^{2} \alpha ( {1 + sin \alpha )}^{2} }{16} - \frac{ {cot}^{2} \alpha (1 - {sin \alpha )}^{2} }{16} )}^{2} \\ \\ = > {( {m}^{2} - {n}^{2}) }^{2} = \frac{ {cot}^{4} \alpha }{ {16}^{2} } {(1 + {sin}^{2} \alpha + 2sin \alpha }^{} - 1 - {sin}^{2} \alpha + 2sin \alpha ) {}^{2} \\ \\ = > {( {m}^{2} - {n}^{2} )}^{2} = \frac{ {cot}^{4} \alpha }{ {16}^{2} } .(16 {sin}^{2} alpha ) \\ \\ = > {( {m}^{2} - {n}^{2} ) }^{2} = \frac{ {cos}^{4} \alpha . {sin}^{2} \alpha }{ 16.{sin}^{4} \alpha } \\ \\ = > {( {m}^{2} - {n}^{2}) }^{2} = \frac{ {cos}^{4} \alpha }{16. {sin}^{2} \alpha } \: \: \: \: ...[Eq _{2}]$
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BY Eq(1) And Eq(2)

$= > {( {m}^{2} - {n}^{2}) }^{2} = mn$
__________________"[PROVED]"

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