How do you simplify #(cos^2ysin(y+pi/2))/(sin^2ysin(pi/2-y)#?

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Answer 1 · 2016-09-30T18:40:34

#cot^2y.#

We know that #sin(A+B)=sinAcosB+sinBcosA#

Using the above formula, we expand,

#sin(y+pi/2)=sinycos(pi/2)+cosysin(pi/2)#,

#=0*siny+1*cosy=cosy#

Similarly, #sin (pi/2-y)=cosy.#

Hence, the exp.#=(cos^2y*cosy)/(sin^2y*cosy)#

#=cos^2y/sin^2y=cot^2y#.