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# Tutor profile: Eidan M.

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Eidan M.
Tutor for 4 years in topics in Math
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## Questions

### Subject:Linear Algebra

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Question:

Show that if two matrices $$U, V$$ are orthogonal then their product $$UV$$ is orthogonal.

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Eidan M.
Answer:

We know that a matrix is orthogonal if and only if the transpose is its own inverse. That is that $$MM^{T} = I\Rightarrow M^{T}=M^{-1}$$. So if $$U,V$$ are orthogonal then $$U^{T}=U^{-1}$$ and $$V^{T}=V^{-1}$$. Now it suffices to show that $$(UV)^{-1} = (UV)^{T}$$. By properties of inverses and transposes of matrices, as well as what we know above of $$U$$ and $$V$$, we have that: $$(UV)^{-1}=V^{-1}U^{-1} = V^TU^T = (UV)^T$$ Therefore $$UV$$ is orthogonal.

### Subject:Pre-Calculus

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Question:

Show that the identity $$\tan\theta \sin\theta +\cos\theta = \sec\theta$$ holds.

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Eidan M.
Answer:

We first select a side to work with as working with the whole equality can easily lead to mistakes. It is often easier to start with the more complex side and attempt to simplify towards the simpler one. So we will begin by trying to deduce $$\sec\theta$$ from the L.H.S. (Left hand side). $$\tan\theta\sin\theta + \cos\theta = \frac{\sin\theta}{\cos\theta}\sin\theta + \cos\theta= \frac{\sin^2\theta}{\cos\theta}+\cos\theta$$ Now we combine the fraction through a common denominator $$=\frac{\sin^2\theta}{\cos\theta}+\cos\theta\frac{\cos\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta}+\frac{\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta+\cos^2\theta}{\cos\theta}$$ Finally using the famous identity which derives from Pythagoras Theorem, we know that $$\sin^2\theta + \cos^2\theta = 1$$ So we conclude that $$\tan\theta\sin\theta + \cos\theta =\frac{\sin^2\theta+\cos^2\theta}{\cos\theta}= \frac{1}{\cos\theta} = \sec\theta$$.

### Subject:Calculus

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Question:

$$\int \ln (x) dx$$

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Eidan M.
Answer:

The problem is solved by integration by parts. We use the trick "LIATE" to determine the $$u$$. Since $$\ln(x)$$ is first in the "L" of "LIATE" which stands for "logarithms", then we have: $$\text{Let } u= \ln(x)\text{, }dv = dx\\ \text{So, }du = \frac{1}{x}dx\text{, }v = x$$ Now, through the formula of integration by parts, $$\int u dv = uv - \int v du$$, we substitute with our inputs: $$\int \ln(x) dx = x\ln(x) - \int x \cdot\frac{1}{x}dx = x\ln(x) - \int 1 dx = x\ln(x) - x + C$$

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