xsin³A+ycos³A=sinA.cosA and xsinA=ycosAProve that x²+y²=1

1.	xsin³A+ycos³A=sinA.cosA and xsinA=ycosAProve that x²+y²=1
Answer» We have,xsin3{tex}\\theta{/tex}\xa0+ ycos3{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0(xsin{tex}\\theta{/tex}) sin2{tex}\\theta{/tex}\xa0+ (ycos{tex}\\theta{/tex}) cos2{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0(sin2{tex}\\theta{/tex}) + (x sin{tex}\\theta{/tex}) cos2{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}\xa0[{tex}\\because{/tex}\xa0x sin{tex}\\theta{/tex}\xa0= y cos{tex}\\theta{/tex}]{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0(sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}) = sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x sin{tex}\\theta{/tex}\xa0= sin{tex}\\theta{/tex}\xa0cos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0x = cos{tex}\\theta{/tex}Now, xsin {tex}\\theta{/tex}\xa0= ycos{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}\xa0= y cos{tex}\\theta{/tex}\xa0[{tex}\\because{/tex} x = cos{tex}\\theta{/tex}]{tex}\\Rightarrow{/tex}\xa0y = sin{tex}\\theta{/tex}Hence, x2 + y2 = cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}\xa0= 1

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