### Challenging question

Question 1

Prove that $ \frac{\tan\theta}{1-\cot\theta} + \frac{\cot\theta}{1-\tan\theta} = 1 + \sec\theta\csc\theta $

From LHS (Do not need to show this in answer, I am doing this to explain why we need to multiply by $ \cos\theta - \sin\theta $ later on )

$\begin{aligned}

\frac{\sin\theta}{\cos\theta} \left(\frac{\sin\theta}{\sin\theta-\cos\theta}\right) + \frac{\cos\theta}{\sin\theta} \left(\frac{\cos\theta}{\cos\theta - \sin\theta} \right) &= \frac{-\sin^3 \theta + \cos^3 \theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\

\end{aligned} $

From RHS

$\begin{aligned}

1 + \frac{1}{\sin\theta}\frac{1}{\cos\theta} &= \frac{\sin\theta \cos\theta + 1}{\sin\theta \cos\theta} \\

&= \frac{\sin\theta \cos\theta + 1}{\sin\theta \cos\theta} \left(\frac{\cos\theta - \sin\theta}{\cos\theta - \sin\theta} \right) \qquad \text{clue from LHS}\\

&= \frac{\sin\theta\cos^2\theta - \sin^2\theta\cos\theta + \cos\theta - \sin\theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\

&= \frac{\sin\theta(1-\sin^2\theta) - (1-\cos^2\theta)\cos\theta + \cos\theta - \sin\theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\

&= \frac{-\sin^3 \theta + \cos^3 \theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\

&= \frac{\sin\theta}{\cos\theta} \left(\frac{\sin\theta}{\sin\theta-\cos\theta}\right) + \frac{\cos\theta}{\sin\theta} \left(\frac{\cos\theta}{\cos\theta - \sin\theta} \right) \\

&= \frac{\tan\theta}{1-\cot\theta} + \frac{\cot\theta}{1-\tan\theta}

\end{aligned} $

Note that the next question is not in the O level A Maths syllabus

Question 2

Differentiate with respect to x $\qquad y = \frac{x \sqrt{ x^2 + 1}}{(x + 1)^{\frac{2}{3}}} $

The simpler way is to use logarithmic differentiation

$\begin{aligned}

y &= \frac{x \sqrt{ x^2 + 1}}{(x + 1)^{\frac{2}{3}}} \\

\ln{y} &= \ln{x} + \frac{1}{2} \ln{(x^2 +1)} - \frac{2}{3} \ln{(x+1)} \\

\text{differentiate implicitly with respect to x } \\

\frac{1}{y} \frac{dy}{dx} &= \frac{1}{x} + \frac{1}{2} \frac{1}{x^2 + 1} 2x - \frac{2}{3}\frac{1}{x+1} \\

\end{aligned} $

The other method is to use quotient rule

$\begin{aligned}

\frac{dy}{dx} &=\frac{(x+1)^\frac{2}{3} (x(\frac{1}{2})(x^2+1)^{-\frac{1}{2}}2x + \sqrt{ x^2 + 1}) -x \sqrt{ x^2 + 1}(\frac{2}{3}(x+1)^\frac{-1}{3}) }{(x + 1)^{\frac{4}{3}}}

\end{aligned} $

You would realize that it gets really messy if quotient rule is used. Hence implicit differentiation is recommended.

Note that the next question is not in the O level A Maths syllabus

Question 3

Find an equation of the curve that passes through the point (0, 1) and whose slope at (x, y) is 3xy

Given $\frac{dy}{dx} = 3xy $

$\begin{aligned}

\frac{dy}{dx} &= 3xy \\

\frac{1}{y} \frac{dy}{dx} &= 3x \\

\text{Integrate by seperating variables}

\int \frac{1}{y} dy &= \int 3x dx \\

\ln{y} &= \frac{3x^2}{2} + c \\

\text{Substitute} x=0 \qquad y = 1 \\

c &= -\frac{3}{2} \\

\end{aligned} $

Prove that $ \frac{\tan\theta}{1-\cot\theta} + \frac{\cot\theta}{1-\tan\theta} = 1 + \sec\theta\csc\theta $

From LHS (Do not need to show this in answer, I am doing this to explain why we need to multiply by $ \cos\theta - \sin\theta $ later on )

$\begin{aligned}

\frac{\sin\theta}{\cos\theta} \left(\frac{\sin\theta}{\sin\theta-\cos\theta}\right) + \frac{\cos\theta}{\sin\theta} \left(\frac{\cos\theta}{\cos\theta - \sin\theta} \right) &= \frac{-\sin^3 \theta + \cos^3 \theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\

\end{aligned} $

From RHS

$\begin{aligned}

1 + \frac{1}{\sin\theta}\frac{1}{\cos\theta} &= \frac{\sin\theta \cos\theta + 1}{\sin\theta \cos\theta} \\

&= \frac{\sin\theta \cos\theta + 1}{\sin\theta \cos\theta} \left(\frac{\cos\theta - \sin\theta}{\cos\theta - \sin\theta} \right) \qquad \text{clue from LHS}\\

&= \frac{\sin\theta\cos^2\theta - \sin^2\theta\cos\theta + \cos\theta - \sin\theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\

&= \frac{\sin\theta(1-\sin^2\theta) - (1-\cos^2\theta)\cos\theta + \cos\theta - \sin\theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\

&= \frac{-\sin^3 \theta + \cos^3 \theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\

&= \frac{\sin\theta}{\cos\theta} \left(\frac{\sin\theta}{\sin\theta-\cos\theta}\right) + \frac{\cos\theta}{\sin\theta} \left(\frac{\cos\theta}{\cos\theta - \sin\theta} \right) \\

&= \frac{\tan\theta}{1-\cot\theta} + \frac{\cot\theta}{1-\tan\theta}

\end{aligned} $

Note that the next question is not in the O level A Maths syllabus

Question 2

Differentiate with respect to x $\qquad y = \frac{x \sqrt{ x^2 + 1}}{(x + 1)^{\frac{2}{3}}} $

The simpler way is to use logarithmic differentiation

$\begin{aligned}

y &= \frac{x \sqrt{ x^2 + 1}}{(x + 1)^{\frac{2}{3}}} \\

\ln{y} &= \ln{x} + \frac{1}{2} \ln{(x^2 +1)} - \frac{2}{3} \ln{(x+1)} \\

\text{differentiate implicitly with respect to x } \\

\frac{1}{y} \frac{dy}{dx} &= \frac{1}{x} + \frac{1}{2} \frac{1}{x^2 + 1} 2x - \frac{2}{3}\frac{1}{x+1} \\

\end{aligned} $

The other method is to use quotient rule

$\begin{aligned}

\frac{dy}{dx} &=\frac{(x+1)^\frac{2}{3} (x(\frac{1}{2})(x^2+1)^{-\frac{1}{2}}2x + \sqrt{ x^2 + 1}) -x \sqrt{ x^2 + 1}(\frac{2}{3}(x+1)^\frac{-1}{3}) }{(x + 1)^{\frac{4}{3}}}

\end{aligned} $

You would realize that it gets really messy if quotient rule is used. Hence implicit differentiation is recommended.

Note that the next question is not in the O level A Maths syllabus

Question 3

Find an equation of the curve that passes through the point (0, 1) and whose slope at (x, y) is 3xy

Given $\frac{dy}{dx} = 3xy $

$\begin{aligned}

\frac{dy}{dx} &= 3xy \\

\frac{1}{y} \frac{dy}{dx} &= 3x \\

\text{Integrate by seperating variables}

\int \frac{1}{y} dy &= \int 3x dx \\

\ln{y} &= \frac{3x^2}{2} + c \\

\text{Substitute} x=0 \qquad y = 1 \\

c &= -\frac{3}{2} \\

\end{aligned} $

plzz just make LHS =RHS

ReplyDeletei mean just solve LHS so it is equal to RHS

thanks.

will vote u the best answer for sure

and in the 1st line u wrote the ques. wrong

u wrote 1-secQ but it is 1+secQ

although u have done it corrwct,....