Question

if tan theta+sin theta=m and tan theta-sin theta=n, then find k if (m^2-n^2)^2=kmn
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Answer 1

m²=tan²Φ+sin²Φ+2 tanΦsinΦ

n²=tan²Φ+sin²-2 tanΦsinΦ

m²-n²=4tanΦsinΦ

$4 \sqrt{mn} = 4 \sqrt{(tanΦ + \sinΦ) (tanΦ - sinΦ)}$

$= 4 \sqrt{ (\ \{ \tan }^{2}Φ - { \ \sin}^{2}Φ)$

$= 4 \sqrt{ \frac{ \ { \sin }^{2} Φ}{ { \cos }^{2} Φ } } - \ { \sin }^{2} Φ$

$4 \sinΦ \sqrt{ \ { \sec }^{2} Φ - 1}$

$4 \sqrt{mn} = 4 \: sinΦ \: tanΦ$

=m²-n²