manipulating Trigonometric equations

Question

If $\tan (\theta)=x\tan(\phi)$ and $\sin(\theta)=y\sin(\phi)$ then what is $\cos^2(\theta)$ in terms of $x$ and $y$.

I have tried to use properties such as if $a=b$ and $c=d$ then $ac=bd$ but it always seem to leave a $\cos$ or $\tan$ term. Is there any way to solve in terms of just $y$ and $x?$

Your solution is greatly appreciated

Answer 1

Eliminate $\phi$ by

$$\tan^2\phi=\frac{\sin^2\phi}{1-\sin^2\phi}$$ or

$$\frac{\tan^2\theta}{x^2}=\frac{\sin^2\theta}{y^2-\sin^2\theta}.$$

Then, simplifying

$$x^2\cos^2\theta=y^2-1+\cos^2\theta$$

and

$$\cos^2\theta=\frac{y^2-1}{x^2-1}.$$

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Alternatively, $$\sin\phi=\frac{\sin\theta}y$$ and

$$\cos\phi=\frac{\sin\phi}{\tan\phi}=\frac xy\cos\theta.$$

Hence $$\frac{\sin^2\theta}{y^2}+\frac{x^2\cos^2\theta}{y^2}=1$$

is

$$(x^2-1)\cos^2\theta=y^2-1.$$

Answer 2

$$\tan \theta = x\tan \phi \implies \frac{\sin\theta}{\cos\theta} =\frac{y\sin\phi}{\cos\theta} = x\frac{\sin\phi}{\cos\phi} \implies \color{red}{\cos\phi = \frac{x}{y}\cos\theta}$$

Also, $$\color{red}{\sin\phi = \frac{1}{y}\sin\theta}$$

$$\cos^2\phi+\sin^2\phi = 1 = \frac{x^2}{y^2}\cos^2\theta + \frac{1}{y^2}\sin^2\theta$$

Now use the fact that $\color{blue}{\sin^2\theta = 1-\cos^2\theta}$ and complete.

Answer 3

As $\csc^2\phi-\cot^2\phi=1$

$$\left(\dfrac y{\sin\theta}\right)^2-\left(\dfrac x{\tan\theta}\right)^2=1$$

$$\implies1=\dfrac{y^2}{1-c^2}-\dfrac{x^2}{\dfrac{1-c^2}{c^2}}=\dfrac{y^2}{1-c^2}-\dfrac{c^2x^2}{1-c^2}$$ where $c=\cos\theta$

$$\implies1-c^2=y^2-x^2c^2\iff c^2(1-x^2)=1-y^2$$