Four easy numerical questions

Question

$7^5 \cdot 3^2 \cdot 5 + 3$, is it a composite number?

Find a quadratic polynomial whose zeroes are $3 + \sqrt{5}$ and $3 - \sqrt{5}$

Solve for x and y $\frac x a + \frac y b = 2$ ; $ax - by = a^2 - b^2$

Prove that $sec^2\theta + cosec^2\theta = sec^2\theta*cosec^2\theta$

Answer 1

Too many questions in too few days, and you show no insights, self effort..., Here are some hints:

$$1)\;\;\; 7^5\cdot 3^2\cdot 5+3=3\,(7^5\cdot 3\cdot 5+1)$$

$$(2)\;\;\;(x-(3+\sqrt 5))(x-(3-\sqrt 5))\;--\text{(Note that this polynomial is rational)}$$

$$(3)\;\;\;\frac{x}{a}+\frac{y}{b}=2\Longrightarrow bx+ay=2ab\Longrightarrow\,\text{we have the linear system:}$$

$$\begin{align*}ax-by=&a^2-b^2\\bx+ay=&2ab\end{align*}$$

Since from the given data $\,a,b\neq 0\,$ (why?), we get above $\,x=a\,$

Multiply now the first eq. by $\,a\,$ and the second one by $\,b\,$ and get:

$$\begin{align*}a^2x-aby=&a^3-ab^2\\b^2x+aby=&2ab^2\end{align*}$$

Now sum both eq's and solve for $\,x\,$ :

$$(a^2+b^2)x=a^3+ab^2=a(a^2+b^2)$$.

Since from the given data $\,a,b\neq 0\,$ (why?), we get above $\,x=a\,$ . Substitute now in either equation and get $\,y\,$ .

$$\sec^2t+\csc^2t=\frac{1}{\cos^2t}+\frac{1}{\sin^2 t}=\frac{\sin^2t+\cos^2t}{\sin^2t\,\cos^2t}$$

Now just remember the trigonometric Pythagoras Theorem $\,\cos^2x+\sin^2x=1\,$ and you get that the above equals what you want.