Trigonometry in projectile motion

Question

I initially posted this question on Physics SE but got no responses probably because it's more related to maths than physics.

A plane surface makes an angle $\bf X$ with the horizontal. From the bottom of this inclined plane, a bullet is fired with velocity $\bf v$. Find the maximum possible range of the bullet on the inclined plane.

Now to start, I assumed the projectile is fired at an angle $\bf Y$ with the inclined surface. The $\bf x$-axis I'm taking is parallel to inclined surface and $\bf y$-axis is perpendicular to it.

So, initial velocity and acceleration in $\bf X$ direction is $\bf v cosY$ and $\bf - g sinX$ and in $\bf Y$direction is $\bf v sinY$ and $\bf - g cos Y$.

I found out the time period of the motion to be $$\bf \frac{(2v sinY)}{(g cosX)}$$ So Range in X which I simplified to $$\bf \frac{2 v^2sin^2Y (\tan Y - \tan X)}{g cosX}$$.

However I have to get the max range only in terms of $\bf v, g and X$. So I have to eliminate $\bf Y$ from the equation which I am not able to do. Please help me with that.

Answer 1

You formed the following equations:

$$u_x=u\cos y;u_y=u\sin y\\a_x=g\sin x;a_y=g\cos x\\T=2\frac{u_y}{a_y}=\frac{2u\sin y}{g\cos x}$$

I form $$R=u_xt-\frac12a_xt^2=u\cos y\left(\frac{2u\sin y}{g\cos x}\right)-\frac12g\sin x\left(\frac{2u\sin y}{g\cos x}\right)^2\\=\frac{2u^2\sin^2y}{g\cos x}(\cot x-\tan x)$$

You formed wrong equation, try again after correcting your equation, this is the most I should help you.