Question

cosec (90-theta) -sin(90-theta) ( cosec theta - sin theta) (tan theta+cot theta) =1​

Answer 1

Given :

cosec(90- x)-sin(90- x)( cosecx - sinx)(tanx + cotx) = 1

Taking L.H.S :

= cosec(90- x)-sin(90- x)( cosecx - sinx)(tanx + cotx)

= (secx - cosx)( 1/sinx - sinx ) ( sinx/cosx + cosx/sinx )

= ( 1/cosx - cosx )( 1 - sin²x / sinx ) ( sin²x + cos²x / cosx.sinx )

= (1 - cos²x / cosx) ( cos²x / sinx ) ( 1 / sinx. cosx )

= ( sin²x.cos²x.1 ) / sin²x.cos²x

= 1

= R.H.S

Answer 2

= cosec (9090-theta)-sin (90-theta )(cos theta-sintheta)=1

=(sec theta-cos theta)(1/sin theta)(sintheta/cos theta+cos theta/sin theta)

=(1/cos theta-cos theta) (1-sin^theta)(sin^theta+cos^theta)

=(1-cos^theta/cos theta)(cos^theta/sintheta)(1/sin theta×vos theta)

=1