<s>[INST] 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 $2\beta$ 

 [/INST] Have:
$\csc A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Now:
$(\csc A-\sin a)(\sec A-\cos A)=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)=(\frac{1-\sin^2{A}}{\sin a})(\frac{1-\cos^2{a}}{\cos A})=\frac{\cos^2 A}{\sin A}\cdot\frac{\sin^2 A}{\cos A}=\cos A\sin A=\frac{\cos A\sin A}{1}=\frac{\cos A\sin A}{\cos^2 A+\sin^2 A}=\frac{\frac{\cos A\sin A}{\cos A\sin A}}{\frac{\cos^2 A+\sin^2 A}{\cos A\sin A}}=\frac{1}{\frac{\cos^2 A}{\cos A\sin A}+\frac{\sin^2 A}{\cos A\sin A}}=\frac{1}{\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}}=\frac{1}{\cot A+\tan A}$
</s>