Find value of $(a^2-1) \tan^2 A + (1-b^2) \tan^2 B$

Question

Find value of $(a^2-1) \tan^2 A + (1-b^2) \tan^2 B$

given $\sin A =a \cos B$ and $\cos A =b \sin B$.

I squared and added conditions to get $a^2 \cos^2 B + b^2 \sin^2 B = 1$. How do i proceed? Thanks

Answer 1

$a=\dfrac{\sin A}{\cos B}$ and $b=\dfrac{\cos A}{\sin B}$ then $$(a^2-1) \tan^2 A + (1-b^2) \tan^2 B=\frac{\sin^2A-\cos^2B}{\cos^2A\cos^2B}=\tan^2 A\tan^2 B-1=\frac{a^2}{b^2}-1$$ because $$\frac{a}{b}=\frac{\dfrac{\sin A}{\cos B}}{\dfrac{\cos A}{\sin B}}=\tan A\tan B$$

Answer 2

HINT:

$$a^2 \cos^2 B + b^2 \sin^2 B =1= \cos^2 B + \sin^2 B$$

$$\iff(a^2-1)\cos^2B=(1-b^2)\sin^2B$$

$$\iff\tan^2B=\dfrac{a^2-1}{1-b^2}$$

Now $\tan A=\cdots=\dfrac{a\cos B}{b\sin B}=\dfrac a{b\tan B}=\cdots$