Question

Prove the following identities:

$se{c}^{4}A(1-si{n}^{4}A)-2ta{n}^{2}A=1$

Answer 1

$LHS=se{c}^{4}A(1-si{n}^{4}A)-2ta{n}^{2}A$

$=se{c}^{4}A-si{n}^{4}Ase{c}^{4}A-2ta{n}^{2}A$

$=se{c}^{4}A-si{n}^{4}A\times \frac{1}{co{s}^{4}A}-2ta{n}^{2}A$

$=se{c}^{4}A-ta{n}^{4}A-2ta{n}^{2}A$

$=\left[\right(se{c}^{2}A{)}^2-\left(ta{n}^{2}A{)}^2\right]-2ta{n}^{2}A$

$=(se{c}^{2}A-ta{n}^{2}A)(se{c}^{2}A+ta{n}^{2}A)-2ta{n}^{2}A$

$=(se{c}^{2}A-ta{n}^{2}A)(se{c}^{2}A+ta{n}^{2}A)-2ta{n}^{2}A$

$(Sincese{c}^{2}A-ta{n}^{2}A=1$)

$=se{c}^{2}A+ta{n}^{2}A-2ta{n}^{2}A$

$=se{c}^{2}A-ta{n}^{2}A=1=LHS$