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One of the trig identities state that \(\tan A = \frac{\sin A}{\cos A}\) (you could just prove it by using sohcahtoa), so that's what we need to solve for. To do that, divide both sides by cos A and then by 2 to get:

\(\frac{\sin A}{\cos A} = \boxed{\frac{3}{2}}\), which is our final answer.

If \(5-i\sqrt[4]{3}\) is a root, then we can set \(5+i\sqrt[4]{3}\) to also be a root, since it is the conjugate of the other root and will multiply to cancel out the imaginary term. It would look something like this:

\((x-(5-i\sqrt[4]{3}))(x-(5+i\sqrt[4]{3}))\\=x^2-10x+25+\sqrt{3}\\=(x-5)^2+\sqrt{3}\)

Again, we could use the same conjugate trick as before to cancel out the radical:

\(((x-5)^2+\sqrt{3})((x-5)^2-\sqrt{3})\\=(x-5)^4-3\\=x^4 - 20 x^3 + 150 x^2 - 500 x + 625-3\\ =\boxed{x^4 - 20 x^3 + 150 x^2 - 500 x + 622}\)

(note that for the second last step you could use the binomial theorem to expand the terms faster.

Sorry, the previous solution was full of mistakes. here's a better one:

6=2*3, so we need to check if the number is divisible by both 2 and 3.

The first term will be an odd number, the second term will be even, the third will be odd, and the fourth will be even, and odd+even+odd+even = even, so we don't need to check if it's divisible by 2, because it must be for all positive integers.

The sum of the first and second terms mod 3 will always be 0 when n is a positive integer because when n is odd the first term will leave a remainder of 2 when divided by 3 and the second term will leave a remainder of 1 divided by 3, and when n is odd the first term will leave a remainder of 1 when divided by 3 and the second term will leave a remainder of 2 divided by 3. (that's because they are both congruent to 2 mod 3, but the powers of the two are always different in terms of parity). Either way, 2+1=3=0 mod 3, so the sum of the first 2 terms will always divide 3.

The third term will always leave a remainder of 1 when divided by 3, so we need the last term to leave a remainder of 2 when divided by 3. That only happens when n is an even number because n+3 will be odd, and when 14 is taken to a power that is odd, it will always leave a remainder of 2. (The reason behind that is that 14 is congruent to -1 mod 3, which means the remainder alternates between -1 = 3-1 = 2 and 1 mod 3).

That means that only even numbers work, so 49 is still our answer.

6=2*3, so we need to check if the number is divisible by both 2 and 3.

The first term will be an odd number, the second term will be even, the third will be odd, and the fourth will be even, and odd+even+odd+even = even, so we don't need to check if it's divisible by 2, because it must be for all positive integers.

That means we only need to check if it's divisible by 3. Notice that the middle 2 terms will both always leave a remainder of -1 when divided by 3 because 8 is congruent to -1 mod 3 and so is 13. 

Because of that, we only need to check if the sum of the first term and the last term will leave a remainder of 2 when divided by 3, since when that remainder is added to the sum of the remainder of the middle two terms, which is -1 + -1 = -2, the remainder becomes 0 since -2+2 leaves a remainder of 0 when divided by 3.

Since 5 is congruent to (-1) mod 3 and so is 14, n must be even so that the sum of the remainders of the first and last terms is equal to 1+1 = 2, which is congruent to 2 mod 3 (as mentioned before that's what we want). If n is odd, then the sum of the remainder of the first and last terms will be equal to (-1)^(2n+1) + (-1)^(2n+1) = 1 mod 3, which is not what we want.

There are exactly 49 even numbers under 100, which means that the answer is \(\boxed{49}\)

First of all, you should be familiar with the binomial theorem, which is: \(\displaystyle (a+b)^n = \sum_{k=0}^n {n\choose k}a^kb^{n-k}\). In this case, \(a=\sqrt{x}, b=\frac{5}{x}\), and \(n=9\). The value of k that is going to produce the constant term is 6, because the x term in a becomes x^3 and the x term in b becomes 1/x^3, and when multiplied it cancels out. That constant term will look like this:

\({9 \choose 6}(\sqrt{x})^6(\frac{5}{x})^3={9 \choose 6}x^3(\frac{125}{x^3})={9 \choose 6}125=84\cdot125=\boxed{10500}\)

1.

\(\frac{1}{\csc x + \cot x}\\ =\frac{1}{\frac{1}{\sin x} + \frac{\cos x }{\sin x}}\\ =\frac{1}{\frac{1+\cos x}{\sin x}}\\ =\frac{\sin x}{1+ \cos x}\\ = \frac{(\sin x)(1- \cos x)}{(1+\cos x)(1- \cos x)}\\ = \frac{(\sin x)(1- \cos x)}{1-(\cos x)^2}\)

by the Pythagorean identity, \((\sin x)^2 + (\cos x)^2 = 1\) (you could prove using Pythagorean's theorem), which means that \(1-(\cos x)^2=(\sin x)^2\). Substitute that back in:

\(\frac{(\sin x)(1-\cos x)}{(\sin x)(\sin(x)}=\frac{1-\cos x}{\sin x}\) and we are done.

2.

\((\csc x)^2(\sec x-1)\\ =(\frac{1}{(\sin x)^2})(\frac{1}{\cos x}-1)\)

Use the Pythagorean identity again:

\((\frac{1}{(1+\cos x)(1-\cos x)})(\frac{1-\cos x}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\frac{1}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\sec x)\\ =\frac{\sec x}{1+\cos x}\)

and we are done. 

Let a and b be real numbers, such that a is the real part of z, and b is the imaginary part. Therefore,

\(2(a+bi)-3i(a-bi)=7-2i\\ 2a+2bi-3ai-3b=7-2i\\ (2a-3b)+(2b-3a)i=7-2i\)

Notice the real part of the right-hand side is 2a-3b, and the real part of the left-hand side is 7, so \(2a-3b=7\). Similarly, the imaginary part of the right-hand side is 2b-3a, and the imaginary part of the left-hand side is -2, so that means that \(-3a+2b=-2\). Now it's just a system of 2 linear equations, and can be easily solved with elimination:

\(6a-9b=21\\-6a+4b=-4\\-5b=17\\b=-\frac{17}{5}\\-3a+2(-\frac{17}{5})=-2\\-3a-\frac{34}{5}=-2\\ -3a=\frac{24}{5}\\a=-\frac{8}{5}\)

Therefore, the complex number z is equal to \(\boxed{-\frac{8}{5}-\frac{17}{5}i}\)

When you get to 2x^2 + 9x + 5 = 0, you could just use Vieta's formulas. The sum \(r+s=-\frac{b}{a}=-\frac{9}{2}\), which means that \((r+s)^2=r^2+2rs+s^2=(-\frac{9}{2})^2=\frac{81}{4}\). The product \(rs=\frac{c}{a}=\frac{5}{2}\), which means \(2rs=5\). Notice that \(r^2+2rs+s^2-2rs=r^2+s^2=\frac{81}{4}-\frac{20}{4}=\boxed{\frac{61}{4}}\)

72 = 8*9, so the number 541G5072H must be divisible by both 8 and 9.

The divisibility rule for 8 is that if the last 3 digits divide 8, then the whole number divides 8. Notice that if you plug in 0 for H, it divides 8 (since 720 = 8*9*10). The only other possibility for H is 8 since no other digits work (you could deduce this by logic).

The divisibility rule for 9 is that if the sum of the digits of that number divides 9, then the whole number divides 9. Notice that we don't need to test out the possibilities of G if H is equal to 0, because the product of GH will be equal to zero (you still could test them out if you wanted to, but it's irrelevant). That means we will only need to test out the possibilities of G if H is equal to 8.

That means that the sum \(5+4+1+G+5+0+7+2+8=32+G\) must divide 9. The only possibility for G is 4, since 32+4=36, which divides 9. That means that \(G=4\) and \(H=8\), and their product is \(\boxed{32}\), which is our answer.

If you add the first 4 equations:

\(p+q+r+pz+py+pz+qx+qy+qz+rx+ry+rz=x+y+z+1\)

substitute x+y+z for -3:

\(p+q+r+pz+py+pz+qx+qy+qz+rx+ry+rz=-2\\(p+q+r)+p(x+y+z)+q(x+y+z)+r(x+y+z)=-2\\ (p+q+r)+(p+q+r)(x+y+z)=-2\\(p+q+r)(1+(x+y+z))=-2\)

substitute x+y+z for -3 again:

\((p+q+r)(-2)=-2\\\boxed{p+q+r=1}\)