Question

(1+(cosecA tanB)^2)/(1+(cosecC tanB)^2)=(1+(cotA sinB)^2)/(1+(cotC sinB)^2)

$\frac{1+(cosecA tanB)^2}{1+(cosecC tanB)^2} =\frac{1+(cotA sinB)^2}{1+(cotC sinB)^2}$)

Answer 1

Answer:

LHS  

= [ 1 + ( csc a tan b )² ] / [ 1 + ( csc c tan b )² ]

= ( 1 + csc² a tan² b ) / ( 1 + csc² c tan² b )

= { tan² b [ ( 1 / tan² b ) + csc² a ] } / { tan² b [ ( 1 / tan² b ) + csc² c ] }

= ( cot² b + csc² a ) / ( cot² b + csc² c )

= [ ( csc² b - 1 ) + ( 1 + cot² a ) ] / [ ( csc² b - 1 ) + ( 1 + cot² c ) ]

= ( csc² b + cot² a ) / ( csc² b + cot² c )  

= { csc² b [ 1 + ( cot² a / csc² b ) ] } / { csc² b [ 1 + ( cot² c / csc² b ) ] }

= ( 1 + cot² a sin² b ) / ( 1 + cot² c sin² b )

= [ 1 + ( cot a sin b )² ] / [ 1 + ( cot c sin b )² ]

= RHS

note: here csc is cosec

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