Question Number 189533 by mustafazaheen last updated on 18/Mar/23 | ||
$${when}\:\:{a}+{b}=\mathrm{60}^{°} \:\:\:\:\:{find}\:\:\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)}=? \\ $$ | ||
Answered by som(math1967) last updated on 18/Mar/23 | ||
$$\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)} \\ $$$$=\frac{{cos}\left({a}+{b}\right){cos}\left({a}−{b}\right)}{{cos}\left({a}−{b}\right)} \\ $$$$={cos}\left({a}+{b}\right)={cos}\mathrm{60}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${if}\:\left({a}−{b}\right)\neq\mathrm{90} \\ $$ | ||