Question #bb3cf

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Answer 1 · 2017-10-25T20:57:36

$"See below"$ .

$x = a /(cos^3 theta)$
$y = b sin^3theta/cos^3theta$ then

$x/y = a/b sin^3 theta$ and then

$theta =arcsin( root(3)((bx)/(ay)))$

Answer 2 · 2017-10-26T02:30:34

We have

$sec^3theta=x$

$=>sec^2theta=x^(2/3)......【1】$

And

$tan^3theta=y$

$=>tan^2theta=y^(2/3)......【2】$

So subtracting [2] from [1] we get

$x^(2/3)-y^(2/3)=sec^2theta-tan^2theta$

$=>x^(2/3)-y^(2/3)=1$