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Solving $\csc <!--\; \; -->(x)+\cot <!--\; \; -->(x)=p$

inverse laplace transform of $\frac{s^3}{(s^2+4{)}^2}$

Expand $(\tan <!--\; \; -->x-<!--\; -\; -->\sqrt{3})(3\tan <!--\; \; -->x+\sqrt{3})$

How to find $\sin <!--\; \; -->(-<!--\; -\; -->\frac{11\mathrm{\pi <!--\; \pi \; -->}}{12})$

Establish the identity:
${\displaystyle \frac{\cot <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\sec <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}=\sec <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\cot <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$

Prove that ${\cos }^6<!--\; \; -->\mathrm{\beta <!--\; \beta \; -->}-<!--\; -\; -->{\sin }^6<!--\; \; -->\mathrm{\beta <!--\; \beta \; -->}=\frac{1}{16}(15\cos <!--\; \; -->2\mathrm{\beta <!--\; \beta \; -->}+\cos <!--\; \; -->6\mathrm{\beta <!--\; \beta \; -->})$

Prove that:
${\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{\sin }^4<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{1}{32}(\cos <!--\; \; -->6\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\cos <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}+2-<!--\; -\; -->2\cos <!--\; \; -->4\mathrm{\theta <!--\; \theta \; -->})$

a, b, c are three complex numbers such that roots of $P(x)=x^3+a{x}^2+bx+c$ lie on a unit circle prove that roots of $Q(x)=x^3+|a|x^2+|b|x+|c|$ lie on unit circle too.

Find all pairs of integers $(x,m)$ for which
$\sqrt[3]{\sqrt[3]{x-<!--\; -\; -->2}+m}+\sqrt[3]{-<!--\; -\; -->\sqrt[3]{x-<!--\; -\; -->2}+m}=2$

Prove formula for $\int <!--\; \int \; -->\frac{dx}{(1+x^2{)}^n}$

With a standard 52 deck (13 types and 4 suites), what is the probability that a poker hand contains cards of five different types? A poker hand is a set of 5 cards.
The obvious answer is A / B where A is the number of poker hands that have 5 different types of cards and B is the total number of poker hands which is C(52,5).
Right now I think A = C(13,5) since there are 13 different types of cards I choose 5 from them but I am unsure if A = 4 * C(13,5) instead, where 4 is the different suites.

How is this trigonometric identify proven?
When researching answers to prove my identity
$\frac{\cos <!--\; \; -->y}{1-<!--\; -\; -->\sin <!--\; \; -->y}$
I get to the point of
$\frac{1+\sin <!--\; \; -->y}{\cos <!--\; \; -->y}$
Then it jumps straight to
$\sec <!--\; \; -->y+\tan <!--\; \; -->y$
How do you prove that identity?

How can we bound $\frac{\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})}{|\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})|+|\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})|}$

Solving ${\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=1$

Show trigonometric identity $\frac{1-<!--\; -\; -->\tan <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}}{1+\tan <!--\; \; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}}=\sec <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$

the following formula can be found in the literature: $|\sin <!--\; \; -->(z){|}^2=\sin <!--\; \; -->(x{)}^2+\sinh <!--\; \; -->(y{)}^2$, z=x+iy; $x,y\in <!--\; \in \; -->\mathbb{R}$.
I am wondering if there is a similar formular in polar coordinates, i.e. $|\sin <!--\; \; -->(r\cdot <!--\; \cdot \; -->e^{i\mathrm{\theta <!--\; \theta \; -->}})|=...?$

Problem:
${\int <!--\; \int \; -->}_0^{\mathrm{\tau <!--\; \tau \; -->}}\sqrt{2}{\cos }^4<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})+3{\cos }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})+7\mathrm{d}\mathrm{\theta <!--\; \theta \; -->}$

Show that
$f(x)=\underset{k=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{c}_k{e}^{-<!--\; -\; -->ikx}=\frac{2\sinh <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->})}{\mathrm{\pi <!--\; \pi \; -->}}\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{(-<!--\; -\; -->1{)}^{k-<!--\; -\; -->1}k}{1+k^2}\sin <!--\; \; -->(kx)$
where ${\displaystyle c_k=\frac{1}{4\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}{e}^{ikx}(e^x-<!--\; -\; -->e^{-<!--\; -\; -->x})dx}$

Derive the following identity:
$\tan <!--\; \; -->(3x)=\frac{3\tan <!--\; \; -->(x)-<!--\; -\; -->{\tan }^3<!--\; \; -->(x)}{1-<!--\; -\; -->3{\tan }^2<!--\; \; -->(x)}$

We have at our disposal the following: the three letters a,b,c , and the five digits 1,2,3,4,5 . We have to form with them all the possible passwords of six (6) characters, under the condition that there must be at least one letter and at least one number in each password. Other than this there are no more restrictions (and, thus, one can repeat at will numbers, letters and etc.)