I've been trying to prove this identity switching between the trigonometric identities but I keep ending up nowhere and eventually get too many repeating identities.
$\endgroup$ 1Prove they are equal. $$ \frac{\cot^2x+1}{\cot^2x} = \sec^2x $$
4 Answers
$\begingroup$we have $$\frac{\cot^2(x)+1}{\cot^2(x)}=\frac{\frac{\cos^2(x)}{\sin^2(x)}+1}{\frac{\cos^2(x)}{\sin^2(x)}}=\frac{\frac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}}{\frac{\cos^2(x)}{\sin^2(x)}}=\frac{1}{\cos^2(x)}$$
$\endgroup$ 1 $\begingroup$If the identities
$$\cot x={1\over\tan x}\qquad\text{and}\qquad1+\tan^2x=\sec^2x$$
are already known to you, then
$${\cot^2x+1\over\cot^2x}=1+{1\over\cot^2x}=1+\tan^2x=\sec^2x$$
$\endgroup$ $\begingroup$You would solve it with the identity $\cot^2x+1 = \csc^2x$, you substitute that, and then simplify the fraction.
$$\frac{\cot^2x+1}{\cot^2x}$$ $$=\frac{\csc^2x}{\cot^2x}$$ $$=\frac{\frac{1}{\sin^2x}}{\frac{\cos^2x}{\sin^2x}}$$ $$=\frac{\sin^2x}{\cos^2x*\sin^2x}$$ $$=\frac{1}{\cos^2x} = \sec^2x$$
$\endgroup$ 1 $\begingroup$It is true that $$x + 1 = \frac{1}{x^{-1}} + 1 = \frac{x^{-1} + 1}{x^{-1}}.$$
Replace the $x$ by $\tan^2 x$, then $$\sec^2 x = \frac{\cot^2 x + 1}{\cot^2 x}.$$
$\endgroup$
