Help with proving identities questions please (1 Viewer)

dabatman · May 6, 2017

Can anyone please help me solve this proof, its from the cambridge 3 unit prelim textbook.

Thanks, ive attached the question.

InteGrand · May 6, 2017

dabatman said:
Can anyone please help me solve this proof, its from the cambridge 3 unit prelim textbook.

Thanks, ive attached the question.

For the first one, multiply top and bottom of the LHS by tan(x). This makes the numerator become tan(x) + 1, and the denominator is tan(x)(1 + tan(x)), so the fraction becomes 1/(tan(x)) = cot(x).

For the second one, multiply top and bottom of the LHS by 1 – sin(α), and remember the difference of two squares and that 1 – sin²(α) = cos²(α).

leehuan · May 6, 2017

$\begin{align*}\frac{1+\cot x}{1+\tan x}&=\frac{\tan x +1}{\tan x + \tan^2 x}\\ &= \frac{1}{\tan x}\times \frac{\tan x + 1}{\tan x + 1}\\ &= \cot x\end{align*}$

First step: Multiply top and bottom by tan(x)
_____________

$\begin{align*}\frac{\cos \alpha}{1+\sin \alpha}&=\frac{\cos\alpha(1-\sin \alpha)}{1-\sin^2\alpha}\\ &= \frac{\cos \alpha(1-\sin \alpha)}{\cos^2\alpha}\\&= \sec^2(1-\sin \alpha)\end{align*}$

dabatman · May 6, 2017

InteGrand said:
For the first one, multiply top and bottom of the LHS by tan(x). This makes the numerator become tan(x) + 1, and the denominator is tan(x)(1 + tan(x)), so the fraction becomes 1/(tan(x)) = cot(x).

For the second one, multiply top and bottom of the LHS by 1 – sin(α), and remember the difference of two squares and that 1 – sin²(α) = cos²(α).

Thanks

dabatman · May 6, 2017

leehuan said:
$\begin{align*}\frac{1+\cot x}{1+\tan x}&=\frac{\tan x +1}{\tan x + \tan^2 x}\\ &= \frac{1}{\tan x}\times \frac{\tan x + 1}{\tan x + 1}\\ &= \cot x\end{align*}$

First step: Multiply top and bottom by tan(x)
_____________

$\begin{align*}\frac{\cos \alpha}{1+\sin \alpha}&=\frac{\cos\alpha(1-\sin \alpha)}{1-\sin^2\alpha}\\ &= \frac{\cos \alpha(1-\sin \alpha)}{\cos^2\alpha}\\&= \sec^2(1-\sin \alpha)\end{align*}$

Thanks

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