Question

Prove that

$\left(i\right)(si{n}^{2}Aco{s}^{2}B-co{s}^{2}Asi{n}^{2}B)=(si{n}^{2}A-si{n}^{2}B)$

$\left(ii\right)(ta{n}^{2}Ase{c}^{2}B-se{c}^{2}Ata{n}^{2}B)=(ta{n}^{2}A-ta{n}^{2}B)$ [2 MARKS]

Answer 1

Concept: 0.5 Mark each
Application: 0.5 Mark each

We have

(i) LHS $=(si{n}^{2}Aco{s}^{2}B-co{s}^{2}Asi{n}^{2}B)$

$=si{n}^{2}A(1-si{n}^{2}B)-(1-si{n}^{2}A)si{n}^{2}B$

$=si{n}^{2}A-si{n}^{2}B=$ RHS


(ii) LHS $=(ta{n}^{2}Ase{c}^{2}B-se{c}^{2}Ata{n}^{2}B)$

$=ta{n}^{2}A(1+ta{n}^{2}B)-(1+ta{n}^{2}A)ta{n}^{2}B$

$=(ta{n}^{2}A-ta{n}^{2}B)=$ RHS