Question

If $tan\theta +sin\theta =m$ and $tan\theta -sin\theta =n,$ then $\frac{(m^2-n^2{)}^2}{mn}=$___.

• A. 16
• B. 8
• C. 4
• D. 1

Answer 1

The correct option is A

16

$tan\theta +sin\theta =m,tan\theta -sin\theta =n$

$m^2-n^2=(tan\theta +sin\theta {)}^2-(tan\theta -sin\theta {)}^2$

$=4tan\theta sin\theta$

$(m^2-n^2{)}^2=16ta{n}^2\theta si{n}^2\theta$ ...(i)

$mn=(tan\theta +sin\theta )(tan\theta -sin\theta )$

$=ta{n}^2\theta -si{n}^2\theta$

$=\frac{si{n}^2\theta }{co{s}^2\theta }-si{n}^2\theta$

$=si{n}^2\theta (\frac{1}{co{s}^2\theta }-1)$

$=si{n}^2\theta (se{c}^2\theta -1)$

$⟹mn=si{n}^2\theta ta{n}^2\theta$ ....(ii)

Dividing eqn (i) by eqn (ii) we get,

$\frac{(m^2-n^2{)}^2}{mn}=\frac{16ta{n}^2\theta si{n}^2\theta }{si{n}^2\theta ta{n}^2\theta }$

$=16$