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Tutor profile: Anil S.

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Anil S.
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Subject: Trigonometry

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Question:

Prove that $$\left ( 1-cot 22^{o} \right )\left ( 1 - cot 23^{o} \right ) = 2$$

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Anil S.
Answer:

Given Question is $$\left ( 1-cot 22^{o} \right )\left ( 1 - cot 23^{o} \right ) = 2$$ Consider left hand side $$\left ( 1-cot 22^{o} \right )\left ( 1 - cot 23^{o} \right )$$ $$= \left ( 1 - \frac{cos23^{o}}{sin 23^{o}} \right )\left ( 1 - \frac{cos22^{o}}{sin 22^{o}} \right )$$ $$ = \left ( \frac{sin23^{o} - cos23^{o}}{sin23^{o}} \right )\left ( \frac{sin22^{o} - cos22^{o}}{sin22^{o}} \right )$$ Multiply numerator and denominator of both fractions by $$\sqrt{2}$$ $$= \left ( \sqrt{2} \right )\left ( \frac{\frac{1}{\sqrt{2}}sin23^{o}-\frac{1}{\sqrt{2}}cos23^{o}}{sin23^{o}} \right ) .\left ( \sqrt{2} \right )\left ( \frac{\frac{1}{\sqrt{2}}sin22^{o}-\frac{1}{\sqrt{2}}cos22^{o}}{sin22^{o}} \right )$$ Apply $$\frac{1}{\sqrt{2}} = sin 45^{o}$$ and $$\frac{1}{\sqrt{2}} = cos 45^{o}$$ $$= \left ( \sqrt{2} \right )\left ( \sqrt{2} \right )\left ( \frac{sin23^{o}cos45^{o} - cos23^{o}sin45^{o}}{sin23^{o}} \right )\times \left ( \frac{sin22^{o}cos45^{o} - cos22^{o}sin45^{o}}{sin22^{o}} \right )$$ Apply formula $$sinA cosB - cosA sinB = sin(A - B)$$ $$=2\times \left ( \frac{sin(23^{o} - 45^{o})}{sin23^{o}} \right )\times\left ( \frac{sin(22^{o} - 45^{o})}{sin22^{o}} \right )$$ $$=2\times \left ( \frac{sin(-22^{o})}{sin23^{o}} \right )\times \left ( \frac{sin(-23^{o})}{sin22^{o}} \right )$$ Apply formula $$ sin(-\theta ) = - sin\theta$$ $$= 2\times \left ( \frac{-sin22^{o}}{sin23^{o}} \right )\times \left ( \frac{-sin23^{o}}{sin22^{o}} \right )$$ $$= 2\times \frac{sin22^{o}}{sin23^{o}} \times \frac{sin23^{o}}{sin22^{o}}$$ Both fractions cancel each other $$= 2$$ Hence proved

Subject: Calculus

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Question:

Evaluate $$ \int_{0}^{3}\frac{x + 1}{x^{2} + 2x + 5}dx $$

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Anil S.
Answer:

Given $$ \int_{0}^{3}\frac {x + 1}{x^{2} + 2 x + 5} dx $$ This calculus problem contains a fraction in which degree of denominator is $$1$$ more than the degree of numerator. Transform the numerator in terms of derivative of denominator $$x + 1 = A\frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{2} + 2 x + 5\right ) + B$$ $$x + 1 = A\left ( 2 x + 2 \right ) + B$$ ...........$$\left ( eqn 1 \right )$$ $$x + 1 = 2 A x + 2 A + B$$ .............................$$\left ( eqn 2 \right )$$ compare coefficient of $$x$$ both sides to get value of $$A$$ $$1 = 2 A$$ divide both sides by $$2$$ to get the value of $$A$$ $$ A = \frac{1}{2}$$ To get value of $$B$$ compare constant term from both sides of $$\left ( eqn2 \right )$$ $$ 1 = 2 A + B$$ substitute value of $$A = \frac {1} {2}$$ $$1 = 2\times \frac{1}{2} + B$$ $$1 = 1 + B$$ Subtract $$1$$ both sides $$B = 0$$ Substitute value of $$A$$ and $$B$$ in $$\left ( eqn 1 \right )$$ $$x + 1 = \frac{1} {2} \left ( 2 x + 2 \right )$$ Intergral becomes $$\int_{0}^{3}\frac{\frac{1}{2}\left ( 2 x + 2 \right )}{x^{2} + 2 x + 5}dx$$ $$= \frac{1}{2}\int_{0}^{3}\frac{2 x + 2}{x^{2} + 2 x + 5}dx$$ Apply formula $$\int \frac{{f}'(x)}{f(x)}dx = logf(x) + c$$, Base of $$logf(x)$$ is $$10$$ In case of upper and lower limit, constant $$c$$ is eliminated On solving Integral becomes $$= \frac{1}{2}\left | log (x^{2} + 2 x + 5) \right |_{0}^{3}$$ Apply limits $$= \frac{1}{2}\left [ log\left ( 20 \right ) - log\left ( 5 \right ) \right ]$$ apply formula $$log(m) - log(n) = log(\frac{m}{n})$$ $$= \frac{1}{2}log(\frac{20}{5})$$ $$= \frac{1}{2}log\left ( 4 \right )$$ $$= \frac{1}{2}log\left ( 2^{2}\right )$$ Apply formula $$log\left ( m^{n} \right ) = n log m$$ $$= \frac{1}{2}\times 2 log 2$$ $$= log 2$$

Subject: Algebra

TutorMe
Question:

find the vertex of the parabola generated by the function $$ f(x) = x^{2} + 4x + 1 $$

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Anil S.
Answer:

Given function is : $$ f(x) = x^{2} + 4 x + 1 $$ let $$ f(x) = y$$ function can be written as $$ y = x^{2} + 4 x + 1 $$ for vertex, a perfect square is made in right hand side The coefficient of $$x$$ is $$4 $$ and half of $$4$$ is $$2$$ Add the square of $$2$$ both sides function can be written as $$ y + 2^{2} = x^{2} + 4 x + 1 +2^{2} $$ This can be written as $$ y + 4 = x^{2} + 4 x + 1 +4 $$ Now subtract 1 both sides $$ y + 4 -1 = x^{2} + 4 x + 1 +4 - 1 $$ In right side $$1$$ cancelled out, function becomes $$ y + 3 = x^{2} + 4 x + 4 $$ Right side is a perfect square, which can be compared with formula $$\left ( a + b\right )^{2}= a^{2} + 2 a b + b^{2}$$ The function will be written as $$ y + 3 = \left ( x + 2 \right )^{2} $$ This is the equation of a parabola of the type $$ y - k = \left ( x - h \right )^{2}$$, where vertex is $$\left ( h,k \right )$$ On comparing we get $$ k = -3 , h = -2$$ Hence vertex of the given function is $$\left ( -2,-3 \right )$$

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