MHB Don't think there is a solution

[dayalji]

Hi if
cosec(x)/(1-cosec(x) -cosec(x)/(1-cosec(x)=50

show that sec^2(x)=25

 

[Greg]

cosec(x)/(1-cosec(x) -cosec(x)/(1-cosec(x)=50

Hi dayalji and welcome to MHB! :D

You have a mismatched parenthesis, so it's difficult to tell what the problem is. Please correct your work.

Also, we ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[dayalji]

Hi dayalji and welcome to MHB! :D

You have a mismatched parenthesis, so it's difficult to tell what the problem is. Please correct your work.

Also, we ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

Hi Greg1313
I have written question again. Problem is the expressions cancel each other out and keep ending with zero.

If $\frac{cosec(x)}{(1-cosec(x)} - \frac{cosec(x)}{(1-cosec(x)} =50 $

show that $sec^2(x)=25$

 

[Greg]

The left hand side of the equation you give above is identically 0, not 50.

 

[soroban]

I think I've found the typo . . .

\text{If }\dfrac{\csc}{1+\csc x} - \frac{\csc x}{1-\csc x} \:=\:50

\text{show that } \sec^2(x)\,=\,25

\begin{array}{cccc}<br /> \text{We have:} &amp; \dfrac{csc x}{1+\csc x} - \dfrac{csc x}{1 - \csc x} &amp;=&amp;50\\ \\<br /> &amp; \dfrac{\csc x(1-\csc x) - \csc x)1+\csc x)}{(1+\csc x)(1 - \csc x)} &amp;=&amp; 50\\ \\<br /> <br /> &amp; \dfrac{-2\csc^2x}{1-\csc^2x} &amp;=&amp; 50\\ \\<br /> <br /> &amp; \dfrac{\frac{-2}{\sin^2x}}{1 - \frac{1}{\sin^2x}} &amp;=&amp; 50 \\ \\<br /> <br /> &amp; \dfrac{-2}{\sin^2x-1} &amp;=&amp; 50\\ \\<br /> <br /> &amp; \dfrac{-2}{-\cos^2x} &amp;-&amp; 50 \\ \\<br /> &amp;2\sec^2x &amp;=&amp; 50 \\ \\<br /> &amp; \sec^2x &amp;=&amp; 25<br /> \end{array}

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