Question

Prove that $ta{n}^{2}a-ta{n}^{2}b=\frac{si{n}^{2}a-si{n}^{2}b}{co{s}^{2}a.co{s}^{2}b}$.

Answer 1

$ta{n}^{2}a-ta{n}^{2}b=\frac{si{n}^{2}a-si{n}^{2}b}{co{s}^{2}a.co{s}^{2}b}$

LHS $ta{n}^{2}a-ta{n}^{2}b$$=\frac{si{n}^{2}a}{co{s}^{2}a}-\frac{si{n}^{2}b}{co{s}^{2}b}$ $=\frac{si{n}^{2}aco{s}^{2}b}{co{s}^{2}a.co{s}^{2}b}-si{n}^{2}bco{s}^{2}a$

$co{s}^{2}b=1-si{n}^{2}b$

$co{s}^{2}b=1-si{n}^{2}b$

Therefore, $\frac{si{n}^{2}a(1-si{n}^{2}b)-si{n}^{2}b(1-si{n}^{2}a)}{co{s}^{2}a.co{s}^{2}b}$

$=\frac{si{n}^{2}a-si{n}^{2}b}{co{s}^{2}a.co{s}^{2}b}$ = RHS