Question

prove that:1/sin2A-1/sin2B=cos2A-cos2B/sin2A*sin2B​

Answer 1

Answer:

$\frac{1}{ { \sin(a) }^{2} } - \frac{1}{ { \sin(b) }^{2} } = \frac{ { \cos(a) }^{2} - { \cos(b) }^{2} }{ { \sin(a) }^{2} . { \sin(b) }^{2} }$

taking LHS

$= \frac{ { \sin(b) ^{2}- { \sin(a) }^{2} } }{ { \sin(a) }^{2} . { \sin(b) }^{2} }$

$= \frac{1 - { \cos(b) }^{2} - (1 - { \cos(a) }^{2} ) }{ { \sin(a) }^{2}. { \sin(b) }^{2} }$

$= \frac{1 - { \cos(b) }^{2} - 1 + { \cos(a) }^{2} }{ { \sin(a) }^{2}. { \sin(b) }^{2} }$

$= \frac{ { \cos(a) }^{2} - { \cos(b) }^{2} }{ { \sin(a) }^{2}. { \sin(b) }^{2} }$

LHS=RHS

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Answer 2

Answer:

damalu dymelu dandanakka nakka nakka vendu tanindathu kada mathsku vanajatha pidu