Question

If $tan\theta +sin\theta =m$ and $tan\theta -sin\theta =n$

Show that $m^2-n^2=4\sqrt{mn}$ [4 MARKS]

Answer 1

Concept: 1 Mark
Substitution: 1 Mark
Application: 2 Marks

$m^2-n^2=(m+n)(m-n)$

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

$=\left(2tan\theta \right)\left(2sin\theta \right)$

$=4tan\theta sin\theta$ ......... (1)

$4\sqrt{mn}=4\sqrt{(tan\theta +sin\theta )(tan\theta -sin\theta )}$

$=4\sqrt{ta{n}^2\theta -si{n}^2\theta }$

$=4\sqrt{\frac{si{n}^2\theta }{co{s}^2\theta }-si{n}^2\theta }$

$=4\sqrt{\frac{si{n}^2\theta -si{n}^2\theta co{s}^2\theta }{co{s}^2\theta }}$

$=4\sqrt{\frac{si{n}^2\theta (1-co{s}^2\theta )}{co{s}^2\theta }}$

$=4tan\theta sin\theta$ .......... (2)

From (1) and (2)

$m^2-n^2=4\sqrt{mn}$