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Question
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
Solution
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
LHS=`cos^2theta/((1-tan theta))+sin ^3theta/((sin theta - cos theta))`
=`cos^2theta/(1-sintheta/costheta)+sin^3 theta/((sin theta-costheta))`
=`cos^3 theta/((cos theta-sin theta))+ sin ^3 theta/((sintheta-cos theta))`
=`(cos^3theta-sin^3 theta)/((costheta - sin theta))`
=`((cos theta-sintheta)(cos^2 theta+cos theta sin +sin^2theta))/((costheta-sintheta))`
=`(sin^2theta + cos^2 theta + cos theta sin theta)`
=`(1+sin theta cos theta)`
=RHS
Hence, L.H.S = R.H.S.
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
