Math, asked by NITHINGOWDA721, 1 year ago

who is intelligent in maths try to solve the above sum ​

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Answered by mysticd
0

cosA/cosB = m ---( 1 )

cosA/sinB = n ------( 2 )

LHS = (m²+n²)cos²B

= [(cosA/cosB)²+(cosA/sinB)²]cos²B

= [1/cos²B+1/sin²B](cos²Acos²B)

=[(sin²B+cos²B)/(cos²Bsin²B)](cos²Acos²B)

= [(cos²Acos²B)/(cos²Bsin²B)]

= [cos²A/sin²B]

= (cosA/sinB)²

= n²

= RHS

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