The question asks me to prove the identity $\tan ^2x-\sin ^2x=\tan^2 x \sin^2 x$ and use this result to explain why $\tan x>\sin x$ for $0<x<90$
I've proved the identity and I can't explain for the second part. Can anyone explain to me? Thanks a lot.
$\endgroup$ 12 Answers
$\begingroup$Since $\tan^2 x\sin^2 x\gt 0$ in our interval, it follows that $\tan^2 x-\sin^2 x$ is positive. Thus $\tan^2 x\gt \sin^2 x$. Because $\tan x$ and $\sin x$ are positive in our interval, we have $$\tan x=\sqrt{\tan^2 x}\gt \sqrt{\sin^2 x}=\sin x.$$
$\endgroup$ $\begingroup$Alternatively, factor the left side:
$$\tan^2(x) - \sin^2(x) = (\tan x - \sin x)(\tan x + \sin x).$$
As the Right Hand Side (RHS) is positive, this expression must be positive as well. Certainly both $\tan x$ and $\sin x$ are positive in this region, hence $\sin x + \tan x$ is as well. Thus $\tan x - \sin x$ has to be positive, otherwise the LHS would be negative.
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