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Trignometry

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5 Apr 2012 09:25:25 IST
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find the number of point on or inside the circle x^2+y^2=4 that satisfy tan^4x+cot^4x+1=3
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Trigonometry

find the number of point on or inside the circle x^2+y^2=4 that satisfy tan^4x+cot^4x+1=3sin^2y

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Joined: 11 Mar 2012
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5 Apr 2012 09:50:37 IST
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5 Apr 2012 13:41:16 IST
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$tan^4x+cot^4x+1=3sin^22y$

$\Rightarrow (tan^2x-cot^2x)^2+2tan^2xcot^2x+1=3sin^22y$

$(tan^2x-cot^2x)^2+2+1=3sin^22y$

$\Rightarrow (tan^2x-cot^2x)^2=3(sin^22y-1)$

Clearly LHS>=0 and RHS<=0,  this is possible only when LHS=RHS=0

$\Rightarrow tan^2x=cot^2x,\; \; sin^22y=1$

Solving these two separately in the range [-2,2] (as (x,y) lies inside or on the circle) we get

$x=-\frac{\pi }{4},\frac{\pi }{4}\; \; ; \; y=-\frac{\pi }{4},\frac{\pi }{4}$

Required points are  :  $\left (-\frac{\pi }{4},-\frac{\pi }{4} \right )\; ; \; \left (-\frac{\pi }{4},\frac{\pi }{4} \right )\; ;\; \left (\frac{\pi }{4},-\frac{\pi }{4} \right )\; ;\; \left (\frac{\pi }{4},\frac{\pi }{4} \right )$

So we have 4 points which satisfy the condition

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