Jamison Rios

2022-07-02

Prove trigonometric identity and how to have ideas
This is the identity:
${\mathrm{cos}}^{2}\left(a\right)-{\mathrm{cos}}^{2}\left(b\right)=\frac{{\mathrm{tan}}^{2}\left(b\right)-{\mathrm{tan}}^{2}\left(a\right)}{\left(1+{\mathrm{tan}}^{2}\left(a\right)\right)\cdot \left(1+{\mathrm{tan}}^{2}\left(b\right)\right)}$

### Answer & Explanation

earendil666h1

Starting from the Pythagorean Theorem:
$co{s}^{2}x+si{n}^{2}x=1$
Then dividing this by $co{s}^{2}x$ we have:
$1+ta{n}^{2}x=se{c}^{2}x$
Now let's look at the LHS:
$co{s}^{2}a-co{s}^{2}b=co{s}^{2}aco{s}^{2}b\left(\frac{1}{co{s}^{2}b}-\frac{1}{co{s}^{2}a}\right)$
$=\frac{1}{se{c}^{2}ase{c}^{2}b}\left(se{c}^{2}b-se{c}^{2}a\right)$
From earlier we know how to substitute $se{c}^{2}x$ and have
$=\frac{1}{\left(1+ta{n}^{2}a\right)\left(1+ta{n}^{2}b\right)}\left(1+ta{n}^{2}b\right)-\left(1+ta{n}^{2}a\right)$
Or simplified:
$\frac{ta{n}^{2}b-ta{n}^{2}a}{\left(1+ta{n}^{2}a\right)\left(1+ta{n}^{2}b\right)}$
The identities often revert back to P.T. at some point and is often a good place to start.

Do you have a similar question?

Recalculate according to your conditions!