Question

$\cos A=a\cos B,\sin A=b\sin B\Rightarrow (b^2-a^2){\sin }^{2}B=$

• A. $1+a^2$
• B. $2+a^2$
• C. $1-a^2$
• D. $2-a^2$

Answer 1

Answer: (C)

$1-a^2$

From question $a=\frac{\cos A}{\cos B}$ and $b=\frac{\sin A}{\sin B}$

$(b^2-a^2){\sin }^{2}B=(\frac{{\sin }^{2}A}{{\sin }^{2}B}-\frac{{\cos }^{2}A}{{\cos }^{2}B}){\sin }^{2}B$

$=\left(\frac{{\sin }^{2}A{\cos }^{2}B{-\cos }^{2}A{\sin }^{2}B}{{{\cos }^{2}B\sin }^{2}B}\right){\sin }^{2}B$

$=\left(\frac{(1-{\cos }^{2}A){\cos }^{2}B{-\cos }^{2}A(1-{\cos }^{2}B)}{{\cos }^{2}B}\right)$

$=\frac{{\cos }^{2}B-{\cos }^{2}A}{{\cos }^{2}B}=1-\frac{{\cos }^{2}A}{{\cos }^{2}B}=1-a^2$