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 $2\beta$

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You should ask each question separately. Also, how about showing some work? – The Chaz 2.0 Sep 20 '12 at 16:45

$\csc A -\sin A=\frac{1-\sin^2A}{\sin A}=\frac{\cos^2A}{\sin A}$

$\sec A -\cos A=\frac{1-\cos^2A}{\cos A}=\frac{\sin^2A}{\cos A}$

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,

$(\frac y 2 )^2-2(\frac y 2 )-15=0\implies y^2-4y-60=0$

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Have:

$\csc A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Now:

$(\csc A-\sin a)(\sec A-\cos A)=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)=(\frac{1-\sin^2{A}}{\sin a})(\frac{1-\cos^2{a}}{\cos A})=\frac{\cos^2 A}{\sin A}\cdot\frac{\sin^2 A}{\cos A}=\cos A\sin A=\frac{\cos A\sin A}{1}=\frac{\cos A\sin A}{\cos^2 A+\sin^2 A}=\frac{\frac{\cos A\sin A}{\cos A\sin A}}{\frac{\cos^2 A+\sin^2 A}{\cos A\sin A}}=\frac{1}{\frac{\cos^2 A}{\cos A\sin A}+\frac{\sin^2 A}{\cos A\sin A}}=\frac{1}{\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}}=\frac{1}{\cot A+\tan A}$

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