Answers

2014-03-23T10:58:40+05:30
 L.H.S=\frac{ COS^{3}A+ SIN^{3} A }{COS A+SIN A} + \frac{ COS^{3}A- SIN^{3}A  }{COS A-SIN A}

L.H.S= \frac{(COS A-SIN A)( COS^{3}A + SIN^{3}A )+(COS A+SIN A)( COS^{3}A - SIN^{3}A )}{(COS A+SIN A)(COS A-SIN A)}

L.H.S= \frac{2 COS^{4}A-2 SIN^{4}A }{ COS^{2}A- SIN^{2}A }

L.H.S=2 \frac{ ( COS^{2}A) ^{2}- ( SIN^{2}A )^{2}  }{COS^{2}A- SIN^{2}A}

L.H.S=2 \frac{ \frac{ (COS 2A+1)^{2} }{ 2^{2} }- \frac{(1-COS 2A)^{2}}{ 2^{2} }  }{COS^{2}A- SIN^{2}A}

L.H.S= \frac{ \frac{ COS^{2}2A+1+2COS2A-1- COS^{2}2A+2COS2A  }{2} }{COS2A}  

L.H.S= \frac{ \frac{4COS2A}{2} }{COS2A}

L.H.S= \frac{2COS2A}{COS2A}

L.H.S=2

L.H.S=R.H.S
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confusion*
2014-03-23T11:35:01+05:30
A^3+b^3=(a+b)(a^2-ab+b^2)
a^3-b^3=(a-b)(a^2+ab+b^2)
so cos^3A+sin^3A=(cosA+sinA)(cos^2A-cosAsinA+sin^2A)
so [cos^3A+sin^3A]/[cosA+sinA]=1-sinAcosA......................1
sin^2A+cos^2A=1
similarly
cos^A-sin^2A=(cosA-sinA)(cos^2A+sinAcosA+sin^2A)
so
[cos^3A-sin^3A]/[cosA-sinA]=1+sinAcosA.............................2
so adding both equation 
1-sinAcosA+1+sinAcosA
2
LHS=RHS

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