If $$\frac{{{ ext{cos }}\alpha }}{{{ ext{cos }}\beta }} = a$$and $$\frac{{{ ext{sin }}\alpha }}{{{ ext{sin }}\beta }} = b{ ext{,}}$$then the value of $${\sin ^2}\beta $$in terms of a and b is?

Question

If $$\frac{{{	ext{cos }}\alpha }}{{{	ext{cos }}\beta }} = a$$and $$\frac{{{	ext{sin }}\alpha }}{{{	ext{sin }}\beta }} = b{	ext{,}}$$then the value of $${\sin ^2}\beta $$in terms of a and b is?

TuteeHUB · Accepted Answer

$\frac{{{a^2} - 1}}{{{a^2} + {b^2}}}$$

TuteeHUB · Answer

$\frac{{{a^2} + 1}}{{{a^2} - {b^2}}}$$

TuteeHUB · Answer

$\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}$$

TuteeHUB · Answer

$\frac{{{a^2} - 1}}{{{a^2} - {b^2}}}$$

1.	If $$\frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a$$and $$\frac{{{\text{sin }}\alpha }}{{{\text{sin }}\beta }} = b{\text{,}}$$then the value of $${\sin ^2}\beta $$in terms of a and b is?
A.	$\frac{{{a^2} + 1}}{{{a^2} - {b^2}}}$$
B.	$\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}$$
C.	$\frac{{{a^2} - 1}}{{{a^2} - {b^2}}}$$
D.	$\frac{{{a^2} - 1}}{{{a^2} + {b^2}}}$$
Answer» D. $\frac{{{a^2} - 1}}{{{a^2} + {b^2}}}$$

If $$\frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a$$and $$\frac{{{\text{sin }}\alpha }}{{{\text{sin }}\beta }} = b{\text{,}}$$then the value of $${\sin ^2}\beta $$in terms of a and b is?

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