Cos square theta divided by 1-tan theta +sin cube theta divided by sin theta -cos theta = 1+sin theta *cos theta

Answer 1

cos^2 θ /(1 - tanθ) + sin^3 θ/(sinθ - cosθ) = 1 + sinθ cosθ
LS = cos^2 θ /(1 - sinθ/cosθ) + sin^3 θ/(sinθ - cosθ)
= cos^2 θ /((cosθ - sinθ)/cosθ) + sin^3 θ/(sinθ - cosθ)
= cos^3 θ /(cosθ - sinθ) - sin^3 θ/(cosθ - sinθ)
= (cos^3 θ - sin^3 θ)/(cosθ - sinθ) <---- I see the difference of cubes
= (cosθ - sinθ)(cos^2 θ + sinθcosθ + sin^2 θ)/(cosθ - sinθ)
= (cos^2 θ + sinθcosθ + sin^2 θ)
= 1 + sinθcosθ
= RS

wheeww!!

Answer 2

If your question mean.

Prove:

cos² ( θ ) / [ 1 - tan ( θ ) ] + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] = 1 + sin ( θ ) ∙ cos ( θ )

then

cos² ( θ ) / [ cos ( θ ) / cos ( θ ) - sin ( θ ) / cos ( θ ) ] + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] =

cos² ( θ ) / [ cos ( θ ) - sin ( θ ) ] / cos ( θ ) + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] =

cos³ ( θ ) / [ cos ( θ ) - sin ( θ ) ] + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] =

- cos³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] + sin³ ( θ ) / [ sin ( θ ) - cos ( θ ) ] =

[ sin³ ( θ ) - cos³ ( θ ) ] / [ sin ( θ ) - cos ( θ ) ] = 1 + sin ( θ ) ∙ cos ( θ ) =

___________________________
Since:
a³ - b³ = ( a - b ) ∙ ( a² + a ∙ b + b² )
___________________________

[ sin ( θ ) - cos ( θ ) ] ∙ [ sin² ( θ ) + sin ( θ ) ∙ cos ( θ ) + cos ( θ )² ] / [ sin ( θ ) - cos ( θ ) ] =

sin² ( θ ) + sin ( θ ) ∙ cos ( θ ) + cos ( θ )² =

sin² ( θ ) + cos ( θ )² + sin ( θ ) ∙ cos ( θ ) =

1 + sin ( θ ) ∙ cos ( θ )

By the way:

1 + sin ( θ ) ∙ cos ( θ ) = 1 + ( 1 / 2 ) sin ( 2 θ )