How to solve this question in more time efficient way?

Question

Q) if$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}$$ then find $4z^2(x^2+y^2)$
a)$(x^2+y^2)^{3}$
b)$(x^2-y^2)^3$
c)$(x^2-y^2)^2$
d)$(x^2+y^2)^2$

Ans:c

i solved this in a very long way:

$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}=z\tan 2a$$$$\implies x= \frac{z\tan 2a}{\sin a} , y=\frac{z\tan 2a}{\cos a}$$

$$x^2+y^2=\frac{4z^2\tan ^22a}{\sin ^2 2a}=\frac{4z^2}{\cos^2 2a}$$ $$\implies z^2=\cos^2 2a(x^2+y^2)/4\ldots (1)$$

now$$x\sin a= y\cos a\implies -2x^2\sin^2 a=-2y^2\cos^2a$$ adding $x^2$ both sides $$x^2(1-2\sin^2 a)= x^2-2y^2\cos^2 a$$$$=x^2-2y^2+2y^2\sin^2a$$ $$=x^2-y^2-y^2+2y^2\sin^2 a$$$$=x^2-y^2-y^2(1-2\sin^2 a)=x^2-y^2-y^2\cos 2a$$$$=(x^2+y^2)\cos 2a= x^2-y^2\implies \cos 2a= \frac{x^2-y^2}{x^2+y^2}\ldots (2)$$

hence from (1) and (2)

$$4z^2(x^2+y^2)= \left( \frac{x^2-y^2}{x^2+y^2} \right)^2 (x^2+y^2)^2= (x^2-y^2)^2$$

now you can see that i got the answer but there was a log way to find this. This question has been asked in an exam so there is noway its solution could be soo long there got to be some shorter way.

So how could i solve this question is more time efficient way?

Answer 1

How about this?

Since $\tan a=y/x$, we have $$y\cos a=\frac{2z(y/x)}{1-(y/x)^2}\quad\Rightarrow\quad \cos a=\frac{2xz}{x^2-y^2}$$ So, from $1+\tan^2a=1/(\cos^2a)$, we have $$1+\left(\frac yx\right)^2=\left(\frac{x^2-y^2}{2xz}\right)^2,$$ i.e. $$4z^2(x^2+y^2)=(x^2-y^2)^2.$$

Answer 2

$x\sin\alpha=y\cos\alpha\iff\tan\alpha=\dfrac yx$

$\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2xy}{x^2-y^2}$

$$\implies\sin\alpha=\dfrac zx\cdot\dfrac{2xy}{x^2-y^2}=\dfrac{2yz}{x^2-y^2}$$

Use $$\dfrac1{(\sin\alpha)^2}-\dfrac1{(\tan\alpha)^2}=1$$

OR

similarly find $\cos\alpha=?$

Now find $\sin^2\alpha+\cos^2\alpha$