Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$${(\csc^2x+1)^2 \over \csc^2x }+{ (\sec^2x+1)^2 \over\sec^2x} = \tan^2x +\cot^2x +7$$ I began working on the left side of the equation, and the first thing I did was get a common denominator, I can't figure out how to get things to start canceling out.

share|cite|improve this question
Start with $$\tan^2x+1=\sec^2x,\cot^2x+1=\csc^2x$$ – lab bhattacharjee May 3 '14 at 16:31
Can you please confirm the correctness of the present version? – lab bhattacharjee May 3 '14 at 16:41
No the correct equation is ((csc^2x+1)^2 /csc^2x) + ((sec^2x+1)^2)/ sec^2x) = tan^2x + cot^2x +7 – Stella May 3 '14 at 16:45
Please double check my edit to verify correctness – abiessu May 3 '14 at 16:50
Not sure why tag 'Trigonometry' is missing – lab bhattacharjee May 3 '14 at 16:56

$${(\csc^2x+1)^2 \over \csc^2x }=\left(\frac{\csc^2x+1}{\csc x}\right)^2=\left(\csc x+\sin x\right)^2\text{ as }\frac1{\csc x}=\sin x$$

So,we have $$(\csc x+\sin x)^2+(\sec x+\cos x)^2=\csc^2x+2+\sin^2x+\sec^2x+2+\cos^2x$$

Now use $$\sin^2x+\cos^2x=1;\sec^2x=\tan^2x+1;\csc^2x=\cot^2x+1$$

share|cite|improve this answer

Your Answer


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.