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

Will anybody please elaborate that how we can prove that both are equal taking left or right? I have tried every method but couldn't find anything.

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    Are you sure that the question is to prove this and not to solve the equation ? – 2017-01-18
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    its been an hour I am trying to solve this but couldnt,yeah i am sure,the book is opened..only to prove. – 2017-01-18
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    Try search for the proof on Approach0: https://approach0.xyz/search/?q=%24%5Cfrac%7B1%2B%5Ccsc%20x%7D%7B%5Ccos%20x-1%7D%3D%5Cfrac%7B1%2B%5Csin%20x%7D%7B1-%5Csin%20x%7D%24&p=1 – 2017-01-18
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    Try substituting $x=\frac{\pi}{6}$ in the given equation. Both sides of the equation are not equal. Either the question is wrong or you are being asked to solve this equation in which case it does not have any solutions as you will get $\sin(x)\cos(x)=1 \implies \sin(2x)=2$ which is simply not possible as the maximum value of $\sin (\theta)$ is 1 – 2017-01-18
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    Could it be a typo and the $\cos$ should be a $\csc$? – 2017-01-18

2 Answers 2

1

Hint: $$1+\csc(x)=1+\frac{1}{\sin(x)}=\frac{1+\sin(x)}{\sin(x)}$$ see my coment

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    now we have $$\frac{1+\sin(x)}{\sin(x)(\cos(x)-1)}=\frac{1+\sin(x)}{1-\sin(x)}$$ – 2017-01-18
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    I have also tried that hint already but not helpful. – 2017-01-18
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    yeah i have added a coment – 2017-01-18
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    Yeah I have done that too but denominator is messing up. – 2017-01-18
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    if $$1+\sin(x)\ne 0$$ then you will get $$1-\sin(x)=\sin(x)\cos(x)-\sin(x)$$ – 2017-01-18
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    @Dr.SonnhardGraubner That would give $1=\sin(x)\cos(x)$ which is not always true. – 2017-01-18
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    and how they are equal?? – 2017-01-18
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    Do you know any other method except that?Sir – 2017-01-18
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    this is true, assuming the equation is to solve we have $$2=2\sin(x)\cos(x)$$ thus we have $$2=\sin(2x)$$ – 2017-01-18
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    @Dr.SonnhardGraubner The maximum value of the sine of an angle is 1. And we are being asked to prove it. Not to solve it. – 2017-01-18
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    this ca not be proved, except we have $$\sin(x)+1=0$$ in the equation above – 2017-01-18
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    Okay Sir thanks but I am still confuse.. – 2017-01-18
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    why this? probably you have made a typo? – 2017-01-18
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$$\frac{1+\csc x}{\cos x-1}=\frac{1+\frac{1}{\sin x}}{\cos x-1}=\frac{1+\sin x}{\cos x\sin x-\sin x}$$

So if

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

It means that

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

It is true only if

$$\sin x =-1→x=\frac{3\pi}{2}+2k\pi$$

or

$$ →\cos x\sin x=1 → \sin (2x)=2$$

which is not true for any $x$.

So the above equality is not a equivalence but a equation with solution

$$x=\frac{3\pi}{2}+2k\pi$$

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    @Muzamil Ali: is it clear? – 2017-01-18