Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\sin x +1}{\cos x}$$

I tried substituting $\sin^2x+\cos^2x = 1$ but I cannot solve it.

share|improve this question

6 Answers 6

up vote 5 down vote accepted

The above method is really verifying and always quick. Another method to arrive at the answer is by rationalising denominator (mainly when the answer [or RHS] is not known or one is asked to work out only from LHS to RHS):

$$\frac{\sin x - \cos x + 1 }{\sin x + \cos x - 1 }\cdot \frac{\sin x + \cos x + 1}{\sin x + \cos x + 1}$$

$$\frac{ (\sin x + 1)^2 - \cos^2 x }{ 2 \sin x \cos x } $$

$$ \frac{ \sin^2 x + 2 \sin x + 1 - \cos^2 x }{ 2 \sin x \cos x } $$

$$ \frac{ \sin^2 x + 2 \sin x + \sin^2 x + \cos^2 x - \cos^2 x } {2 \sin x \cos x } $$

and the answer follows i.e. $$ \frac{\sin x + 1}{\cos x}. $$

Hope it was helpful.

share|improve this answer
    
well done i say :) –  Praphulla Koushik Dec 12 '13 at 13:47
1  
I wouldn't call this rationalizing the denominator, as there's nothing necessarily rational or irrational about the denominator before or after the initial multiplication, but this technique does mirror the technique for rationalizing denominators: using conjugates. –  Isaac Dec 12 '13 at 16:05

Hint

$$\frac{a}{b}=\frac c d\iff ad=bc$$

share|improve this answer
    
and obviously this is allowed when $(b,d)\not=0$ which is when $x \not=\pi/2 + k\pi$ and $x\not=k\pi$ –  Jekyll Dec 12 '13 at 13:39
    
Exactly what's needed here! +1 –  amWhy Dec 13 '13 at 16:00

$(\sin x- \cos x+1)\cos x = \sin x \cos x -\cos^2 x +\cos x$

$(\sin x+ \cos x-1)(\sin x +1) = \sin^2 x + \sin x \cos x +\cos x -1 = \sin x \cos x +\cos x +(\sin^2 x -1)= \sin x \cos x -\cos^2 x +\cos x$

share|improve this answer

As always, the method that "always" (never say "never" OR "always"...) work is the substitution $ t = \tan \frac{x}{2},$ which makes $\sin(x) = \frac{2x}{1+x^2}, \cos(x)= \frac{1-x^2}{1+x^2},$ which makes an identity like this a mechanical verification.

share|improve this answer

Observe that the Right Hand side $\displaystyle\frac{\sin x+1}{\cos x}=\tan x+\sec x$

So, I want to utilize $\displaystyle\sec^2x-\tan^2x=1$

Dividing the numerator & the denominator by $\cos x,$

$$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\tan x-1+\sec x}{\tan x+1-\sec x}$$

$$=\frac{\tan x+\sec x-(\sec^2x-\tan^2x)}{\tan x+1-\sec x}(\text{ Replacing }1\text{ with } \sec^2x-\tan^2x)$$

$$=\frac{(\sec x+\tan x)-(\sec x+\tan x)(\sec x-\tan x)}{\tan x+1-\sec x}$$

$$=\frac{(\sec x+\tan x)(1-\sec x+\tan x)}{\tan x+1-\sec x}$$

$$=\sec x+\tan x$$

share|improve this answer
    
@dona12, how about this? –  lab bhattacharjee Dec 12 '13 at 14:56

I will start like Timotej, but finish differently.

$$\cos x(\sin x-\cos x+1)=\cos x(1+\sin x)-\cos^2x=\cos x(1+\sin x)-(1-\sin^2x)$$ $$=(1+\sin x)\{\cos x-(1-\sin x)\}$$

$$\implies \cos x(\sin x-\cos x+1)=(1+\sin x)(\sin x+\cos x-1)$$

Now change the sides of $\displaystyle \cos x, \sin x+\cos x-1$


Let me derive some other identities

$(1)\displaystyle\sin x(\sin x-\cos x+1)=\sin^2x+\sin x(1-\cos x)=1-\cos^2x+\sin x(1-\cos x)$

$\displaystyle\implies\sin x(\sin x-\cos x+1)=(1-\cos x)(\sin x+\cos x+1)$

$(2)\displaystyle\sin x(\sin x+\cos x-1)=\sin^2x-\sin x(1-\cos x)=(1-\cos x)(1+\cos x-\sin x)$

$(3)\displaystyle\cos x(\sin x+\cos x-1)=\cos^2x-\cos x(1-\sin x)=(1-\sin x)(1+\sin x-\cos x)$

and so on

share|improve this answer
    
@dona12, another method –  lab bhattacharjee Dec 13 '13 at 3:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.