Trigonometric identity and roots of a polynomial. Prove that 
$$(\operatorname{cosec} A–\sin A) (\sec A–\cos A) = \frac {1}{\tan A + \cot A} $$
Also help me with this question please

If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2–2x–15$ then find a quadratic polynomial whose  series [roots?] are $2\alpha$ and $2\beta$ 

 A: $\csc A -\sin A=\frac{1-\sin^2A}{\sin A}=\frac{\cos^2A}{\sin A}$
$\sec A -\cos A=\frac{1-\cos^2A}{\cos A}=\frac{\sin^2A}{\cos A}$
Multiplying we get, $\sin A \cos A$
Now $\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$
$=\frac{\sin^2A+\cos^2A}{\cos A\sin A}=\frac{1}{\cos A\sin A}$
So, $\alpha+\beta=\frac{2}{1}=2$ and $\alpha\beta=-15$
So, $(y-2\alpha)(y-2\beta)=0\implies y^2-2(\alpha+\beta)y+4\alpha \beta=0$
$\implies y^2-4y-60=0$
Alternatively, we need to find the equation whose roots are double to that of $x^2-2x-15=0$. If $x$ is a root of the given equation, and $y$ be a root of the required equation, then $y=2x\implies x=\frac y 2 .$
Replacing $x$ with $\frac y 2$ in $x^2-2x-15=0$ we get,
$(\frac y 2 )^2-2(\frac y 2 )-15=0\implies y^2-4y-60=0 $
A: Have:
$\csc A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Now:
$(\csc A-\sin a)(\sec A-\cos A)=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)=(\frac{1-\sin^2{A}}{\sin a})(\frac{1-\cos^2{a}}{\cos A})=\frac{\cos^2 A}{\sin A}\cdot\frac{\sin^2 A}{\cos A}=\cos A\sin A=\frac{\cos A\sin A}{1}=\frac{\cos A\sin A}{\cos^2 A+\sin^2 A}=\frac{\frac{\cos A\sin A}{\cos A\sin A}}{\frac{\cos^2 A+\sin^2 A}{\cos A\sin A}}=\frac{1}{\frac{\cos^2 A}{\cos A\sin A}+\frac{\sin^2 A}{\cos A\sin A}}=\frac{1}{\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}}=\frac{1}{\cot A+\tan A}$
