Prove the identity
\(\sec^2x+\sec x\tan x \equiv \dfrac{1}{1-\sin x}\).
Solution
\[
\sec^2x+\sec x\tan x=\frac{1}{\cos^2x}+\frac{\sin x}{\cos^2x}
=\frac{1+\sin x}{\cos^2x}
=\frac{1+\sin x}{1-\sin^2x}
=\frac{1}{1-\sin x}.
\]
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