# Trigonometry $1+\tan{x}=\frac{\sin{(\frac{\pi}{4}+x)}}{\cos{\frac{\pi}{4}}\cos{x}}$

How does one get the following equalities ?

$1+\tan{x}=\frac{\sin{(\frac{\pi}{4}+x)}}{\cos{\frac{\pi}{4}}\cos{x}}=\sqrt{2}\frac{\cos{(\frac{\pi}{4}-x)}}{\cos{x}}$

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Is this homework ? If it is, please include the homework tag. –  Amr Jan 20 '13 at 15:41
No this is self-study. Is there a self-study tag ? –  Charles Jan 20 '13 at 17:12

## 1 Answer

$$1+\tan x=1+\frac{\sin x}{\cos x}=\frac{\cos x+\sin x}{\cos x}\frac{\frac1{\sqrt2}}{\frac1{\sqrt2}}=\frac{\cos x\sin \frac {\pi}{4}+\sin x\cos \frac {\pi}{4}}{\cos \frac{\pi}{4}\cos x}=\frac{\sin\left(x+\frac{\pi}{4}\right)}{\cos \frac{\pi}{4}\cos x}$$ Using the fact that $\cos \frac {\pi}{4}=\sin \frac {\pi}{4}=\frac1{\sqrt2}$

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Thank you very much ! And what about the last part ? –  Charles Jan 20 '13 at 17:13
The last one follows from the fact that $\cos \frac {\pi}{4}=\frac1{\sqrt2}$ and $\sin\left(\frac {\pi}{2}-x\right)=\cos x$ –  Dennis Gulko Jan 20 '13 at 19:50
Thank you very much ! –  Charles Jan 20 '13 at 23:02