Question

If $\sin A=a\cos B$ and $\cos A=b\sin B$ then $(a^2-1){\tan }^{2}A+(1-b^2){\tan }^{2}B$ is equal to $\left(\frac{a^2-b^2}{b^2}\right).$

• A. True
• B. False

Answer 1

The correct option is A True

Given:$\sin A=a\cos B$ and $\cos A=b\sin B$
On squaring and adding, we get
${\sin }^{2}A+{\cos }^{2}A=a^2{\cos }^{2}B+b^2{\sin }^{2}B$
$\Rightarrow 1=a^2{\cos }^{2}B+b^2{\sin }^{2}B$
$\Rightarrow a^2(1-{\sin }^{2}B)+b^2{\sin }^{2}B=1$
$\Rightarrow (a^2-b^2){\sin }^{2}B=a^2-1$
$\Rightarrow {\sin }^{2}B=\frac{a^2-1}{a^2-b^2}$
$\Rightarrow 1=a^2{\cos }^{2}B+b^2{\sin }^{2}B$
$\Rightarrow 1=a^2{\cos }^{2}B+b^2(1-{\cos }^{2}B)$
$\Rightarrow (a^2-b^2){\cos }^{2}B=1-b^2$
$\Rightarrow {\cos }^{2}B=\frac{1-b^2}{a^2-b^2}$
$\Rightarrow {\tan }^{2}B=\frac{{\sin }^{2}B}{{\cos }^{2}B}$
$=\frac{\frac{a^2-1}{a^2-b^2}}{\frac{1-b^2}{a^2-b^2}}$
$=\frac{a^2-1}{1-b^2}$
and ${\tan }^{2}A=\frac{a^2}{b^2}{\cot }^{2}B$
$=\frac{a^2}{b^2}.\frac{1-b^2}{a^2-1}$
$\Rightarrow (a^2-1){\tan }^{2}A+(1-b^2){\tan }^{2}B=\frac{a^2}{b^2}.(1-b^2)+(a^2-1)$
$\Rightarrow (a^2-1){\tan }^{2}A+(1-b^2){\tan }^{2}B=\frac{a^2-b^2}{b^2}$