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Sagot :
I think your question means (sin^(2)x)/(1+cosx), if yes then your answer is:
so the trig identity sin^(2) x = 1 - cos^(2)x , which is the form of expression of a^2 - b^2, which by factoring equals (a-b)(a+b).
So 1-cos^(2)x will become (1-cosx)(1+cosx)
Your final fraction would be [(1-cosx)(1+cosx)]/(1+cosx), where 1+cosx will cancel itself on the numerator and the denominator, thus your answer will be 1-cosx
so the trig identity sin^(2) x = 1 - cos^(2)x , which is the form of expression of a^2 - b^2, which by factoring equals (a-b)(a+b).
So 1-cos^(2)x will become (1-cosx)(1+cosx)
Your final fraction would be [(1-cosx)(1+cosx)]/(1+cosx), where 1+cosx will cancel itself on the numerator and the denominator, thus your answer will be 1-cosx
[tex]\frac { \sin ^{ 2 }{ x } }{ 1+\cos { x } } \\ \\ =\frac { 1-\cos ^{ 2 }{ x } }{ 1+\cos { x } }[/tex]
[tex]\\ \\ =\frac { { 1 }^{ 2 }-\cos ^{ 2 }{ x } }{ \left( 1+\cos { x } \right) } \\ \\ =\frac { \left( 1+\cos { x } \right) \left( 1-\cos { x } \right) }{ \left( 1+\cos { x } \right) } \\ \\ =1-\cos { x } [/tex]
[tex]\\ \\ =\frac { { 1 }^{ 2 }-\cos ^{ 2 }{ x } }{ \left( 1+\cos { x } \right) } \\ \\ =\frac { \left( 1+\cos { x } \right) \left( 1-\cos { x } \right) }{ \left( 1+\cos { x } \right) } \\ \\ =1-\cos { x } [/tex]
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