Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 189533 by mustafazaheen last updated on 18/Mar/23

when  a+b=60^°      find  ((cos^2 a−sin^2 b)/(cos(a−b)))=?

$${when}\:\:{a}+{b}=\mathrm{60}^{°} \:\:\:\:\:{find}\:\:\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)}=? \\ $$

Answered by som(math1967) last updated on 18/Mar/23

((cos^2 a−sin^2 b)/(cos(a−b)))  =((cos(a+b)cos(a−b))/(cos(a−b)))  =cos(a+b)=cos60=(1/2)  if (a−b)≠90

$$\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)} \\ $$$$=\frac{{cos}\left({a}+{b}\right){cos}\left({a}−{b}\right)}{{cos}\left({a}−{b}\right)} \\ $$$$={cos}\left({a}+{b}\right)={cos}\mathrm{60}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${if}\:\left({a}−{b}\right)\neq\mathrm{90} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com