Question

hii friends,
plz solve it.....​

Answer 1

Step-by-step explanation:

We need to use the cos rule of 2θ which is:

$\cos(2θ) = { \cos}^{2}θ - \sin^{2} θ$

Answer 2

Step-by-step explanation:

$\sf In \triangle ABC$

$\sf \angle C = \frac {\pi}{2}$

$\sf Hence\:by\:pythagorus \:theorem$

$\sf {a}^{2} + {b}^{2} = {c}^{2}$

$\sf We\: have \: to \: prove$

$\sf sin ( A - B ) = \frac { {a}^{2} - {b}^{2}}{{c}^{2}}$

$\sf From \: sine \:rule \: we \:know \:that$

$\sf \frac {a}{sinA } = \frac {b}{sinB } = \frac {c}{sinC} = k$

$\sf a = k sinA$

$\sf b = k sinB$

$\sf c = k sinC$

$\sf R.H.S. = \frac { {a}^{2} - {b}^{2}}{{c}^{2}}$

$\sf = \frac { {k}^2 {sin}^{2}A - {k}^{2}{sin}^{2}B}{{k}^{2}{sin}^{2}C}$

$\sf \frac { ( sinA - sinB )(sinA + sinB)}{sin (A+B).sin (A+B)}$

$\sf \frac {2sin(\frac {A + B}{2})cos (\frac{A-B}{2}).2sin(\frac {A - B}{2})cos (\frac {A+B}{2})}{sin (A+B)sin (A+B)}$

$\sf \frac {sin ( A + B) sin ( A - B ) }{sin (A+ B )}$

$\sf sin ( A - B )$

THANKS!!!