trefoniu1

2022-01-28

How do I prove the identities of these questions?
$\frac{1+{\mathrm{tan}}^{2}u}{1-{\mathrm{tan}}^{2}u}=\frac{1}{{\mathrm{cos}}^{2}u-{\mathrm{sin}}^{2}u}$
$\frac{\mathrm{sin}x+\mathrm{sin}\left(5x\right)}{\mathrm{cos}x+\mathrm{cos}\left(5x\right)}=\mathrm{tan}\left(3x\right)$

ul2ph3ojc

Expert

You will need two facts, first the definition of tangent, second main trigonometric identity:
$\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}$
${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$
Using that you may transform (1) as:
$\frac{1+{\mathrm{tan}}^{2}u}{1-{\mathrm{tan}}^{2}u}=\frac{1+\frac{{\mathrm{sin}}^{2}u}{{\mathrm{cos}}^{2}u}}{1-\frac{{\mathrm{sin}}^{2}u}{{\mathrm{cos}}^{2}u}}=$
$=\frac{\frac{{\mathrm{cos}}^{2}u+{\mathrm{sin}}^{2}u}{{\mathrm{cos}}^{2}u}}{\frac{{\mathrm{cos}}^{2}u-{\mathrm{sin}}^{2}u}{{\mathrm{cos}}^{2}u}}=\frac{1}{{\mathrm{cos}}^{2}u}\cdot \frac{{\mathrm{cos}}^{2}u}{{\mathrm{cos}}^{2}u-{\mathrm{sin}}^{2}u}=\frac{1}{{\mathrm{cos}}^{2}u-{\mathrm{sin}}^{2}u}$
The original (2) which is:
$\mathrm{sin}x+\frac{\mathrm{sin}5x}{\mathrm{cos}x+\mathrm{cos}5x}=\mathrm{tan}3x$
is false as shown in another answer (take $x=\frac{\pi }{4}$ for instance). Here is a visualization of LHS and RHS

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