1 - frac{{{{sin }^2}y}}{{1 + cos ,y}} + | vedclass

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Question

(d) The expression can be written as

$\frac{{1 + \cos y - {{\sin }^2}y}}{{1 + \cos y}} + \frac{{(1 - {{\cos }^2}y) - {{\sin }^2}y}}{{\sin y\,(1 - \

Vedclass · Accepted Answer

$\cos \,y$

Vedclass · Answer

(d) The expression can be written as

$\frac{{1 + \cos y - {{\sin }^2}y}}{{1 + \cos y}} + \frac{{(1 - {{\cos }^2}y) - {{\sin }^2}y}}{{\sin y\,(1 - \cos y)}}$

$ = \frac{{\cos y\,(1 + \cos y)}}{{1 + \cos y}} + 0 = \cos y.$

$1 - \frac{{{{\sin }^2}y}}{{1 + \cos \,y}} + \frac{{1 + \cos \,y}}{{\sin \,y}} - \frac{{\sin \,\,y}}{{1 - \cos \,y}} =$

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