Question

If $\frac{\cos \alpha }{\cos \beta }=a,\frac{\sin \alpha }{\sin \beta }=b$, then

• A. ${\sin }^2\beta =\frac{a^2-1}{a^2-b^2}$
• B. ${\cos }^2\beta =\frac{a^2-1}{a^2-b^2}$
• C. ${\cos }^2\alpha =\frac{(b^2-1)a^2}{b^2-a^2}$
• D. ${\sin }^2\alpha =\frac{(b^2-1)a^2}{b^2-a^2}$

Answer 1

Answer: (A), (C)

The correct options are
A ${\sin }^2\beta =\frac{a^2-1}{a^2-b^2}$
C ${\cos }^2\alpha =\frac{(b^2-1)a^2}{b^2-a^2}$

Given, $\frac{\cos \alpha }{\cos \beta }=a,\frac{\sin \alpha }{\sin \beta }=b$

$\Rightarrow {\cos }^2\alpha =a^2{\cos }^2\beta$ and ${\sin }^2\alpha =b^2{\sin }^2\beta$

Adding both, $a^2{\cos }^2\beta +b^2{\sin }^2\beta =1$

$\Rightarrow a^2(1-{\sin }^2\beta )+b^2{\sin }^2\beta =1$

$\Rightarrow {\sin }^2\beta =\frac{a^2-1}{a^2-b^2}\Rightarrow {\cos }^2\beta =1-\frac{a^2-1}{a^2-b^2}=\frac{1-b^2}{a^2-b^2}$

also we have, ${\cos }^2\alpha =a^2{\cos }^2\beta$

$\Rightarrow {\cos }^2\alpha =\frac{a^2(1-b^2)}{a^2-b^2}$

$\Rightarrow {\sin }^2\alpha =1-\frac{a^2(1-b^2)}{a^2-b^2}=\frac{b^2(a^2-1)}{a^2-b^2}$