# Prove the identity ${(\csc^2x+1)^2 \over\csc^2x }+ {(\sec^2x+1)^2\over \sec^2x} = \tan^2x +\cot^2x +7$

$${(\csc^2x+1)^2 \over \csc^2x }+{ (\sec^2x+1)^2 \over\sec^2x} = \tan^2x +\cot^2x +7$$ 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.

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