Show these equations in terms of x and y only - ∅ has to be eliminated. x=sin ∅ +cos ∅ , y=tan ∅ +cot ∅ (hint find x2y)

Question

Can anyone do this question?

Show these equations in terms of x and y only - ∅ has to be eliminated. x=sin ∅ +cos ∅ , y=tan ∅ +cot ∅ (hint find x2y)

Answer 1

By expanding $\left[\sin(\phi)+\cos(\phi)\right]^2$, you'll find that

$$x^2=1+2\sin(\phi)\cos(\phi)$$

Multiplying this by $y=\tan(\phi)+\cot(\phi)$ gives

\begin{align*} x^2 y &=\left[1+2\sin(\phi)\cos(\phi)\right]\left[\tan(\phi)+\cot(\phi)\right]\\ &= \left[1+2\sin(\phi)\cos(\phi)\right]\frac{\sin(\phi)}{\cos(\phi)}+\left[1+2\sin(\phi)\cos(\phi)\right]\frac{\cos(\phi)}{\sin(\phi)}\\ &= \frac{\sin(\phi)}{\cos(\phi)}+2\sin ^{2}(\phi)+\frac{\cos(\phi)}{\sin(\phi)}+2\cos ^{2}(\phi)\\ &= \tan(\phi)+\cot(\phi)+2\\ &= y+2 \end{align*}

so the equation is $x^2 y=y+2$.

Answer 2

Hint:

Let us find how the problem came into being.

$$y=\tan\phi+\cot\phi=\cdots=\dfrac1{\sin\phi\cos\phi}$$

$$x^2=(\sin\phi+\cos\phi)^2=1+2\sin\phi\cos\phi$$

Please compare the two values of $\sin\phi\cos\phi$ to eliminate $\phi$