# Tutor profile: Ikechukwu "Daniel" A.

## Questions

### Subject: Pre-Calculus

Find the quotient of $$\frac{2i+3}{7-6i}$$ and simplify. Then, find the exact modulus of the result.

Multiply both the numerator and denominator by the conjugate of the denominator $$\frac{2i+3}{7-6i} *\frac{7+6i}{7+6i} = \frac{(2i+3)(7+6i)}{85} = \frac{(2i+3)(7+6i)}{85} = \frac{9+32i}{85} $$ Now, we can find the modulus of our result, which is the same as simply finding the distance this complex number is from the origin of the complex plane. $$|\frac{9+32i}{85}| = \sqrt{(\frac{9}{85})^2+(\frac{32}{85})^2} =\sqrt{\frac{13}{85}} = \frac{\sqrt{1105}}{85} $$

### Subject: Trigonometry

Prove that the $$\frac{\csc{\theta}}{\sin{\theta}} - \frac{\cot{\theta}}{\tan{\theta}} = 1$$

To prove this identity, we'll start with the left-hand side. To make things simpler for us, we'll rewrite everything in terms of $$\sin\theta$$ and $$\cos\theta$$. $$ \frac{\csc{\theta}}{\sin{\theta}} - \frac{\cot{\theta}}{\tan{\theta}} = \frac{\frac{1}{\sin\theta}}{\sin\theta} - \frac{\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}}$$ $$= \frac{1}{{\sin}^2\theta} - \frac{\cos^2\theta}{\sin^2\theta}$$ $$= \frac{1-\cos^2\theta}{\sin^2\theta}=\frac{\sin^2\theta}{\sin^2\theta} = 1$$ Therefore this identity is true.

### Subject: Calculus

Determine the indefinite integral of $$\int \frac{x^2-8}{(4x-3)(2x+1)}dx$$.

$$\int \frac{x-8}{(4x-3)(2x+1)}dx$$ We can break up the function that is being integrated into its partial fraction decomposition like so: $$ \frac{x-8}{(4x-3)(2x+1)} = \frac{A}{4x-3} + \frac{B}{2x+1}$$ After doing this, we rewrite this expression so that there are no fractions: $$ x-8 = A(2x+1) + B(4x-3)$$ At this point, we must solve for $$A$$ and $$B$$. If $$x = -\frac{1}{2} $$, then $$ -\frac{1}{2}-8 = A(2(-\frac{1}{2})+1) + B(4(-\frac{1}{2})-3)$$ $$ -\frac{1}{2}-8 = A(-1+1) + B(-2-3)$$ As we can see, $$A$$ will disappear, so we can solve for $$B$$. $$ \frac{-17}{2} = -5B$$ $$ \frac{17}{10} = B$$ Now we can go and solve for $$A$$. If $$x = -\frac{3}{4} $$, then $$ \frac{3}{4}-8 = A(2(\frac{3}{4})+1) + B(4(\frac{3}{4})-3)$$ $$ \frac{3}{4}-8 = A(\frac{6}{4}+1) + B(3-3)$$ This time, $$B$$ disappears, so we can now go about solving for $$A$$. $$ \frac{-29}{4} = \frac{10}{4}A $$ $$ A = \frac{-29}{10} $$ Now that we know both $$A$$ and $$B$$, we can go back to our original integral and solve. $$\int \frac{A}{4x-3} + \frac{B}{2x+1}dx = \int \frac{-29}{(10)(4x-3)}+\frac{17}{(10)(2x+1)}dx $$ $$ = \frac{-29}{40}\ln|4x-3| + \frac{17}{20}\ln|2x+1| + C$$

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