Question

Express $\frac{1 - cos \theta + sin \theta}{1 + cos \theta + sin \theta}$ in terms of tan $(\frac{\theta}{2})$

Answer 1

we know, 1 - cosx = 2sin²(x/2)

so, $1-cos\theta=2sin^2(\theta/2)$

similarly, 1 + cosx = 2cos²(x/2)

so, $1+cos\theta=2cos^2(\theta/2)$

and sin2x = 2sinx . cosx

so, $sin\theta=2sin(\theta/2)cos(\theta/2)$

$\frac{1-cos\theta+sin\theta}{1+cos\theta+sin\theta}\\\\=\frac{2sin^2(\theta/2)+2sin(\theta/2)cos(\theta/2)}{2cos^2(\theta/2)+2sin(\theta/2)cos(\theta/2)}\\\\=\frac{2sin(\theta/2)\{sin(\theta/2)+cos(\theta/2)\}}{2cos(\theta/2)\{cos(\theta/2)+sin(\theta/2)\}}\\\\=\frac{sin(\theta/2)}{cos(\theta/2)}\\\\=tan(\theta/2)$

Answer 2

HELLO DEAR,



GIVEN:- $\bold{\frac{1 - cos\Theta + sin\Theta}{1 + cos\Theta + sin\Theta}}$

$\bold{\implies \frac{(1 - cos\Theta) + 2sin\Theta/2.cos\Theta/2}{(1 + cos\Theta) + 2sin\Theta/2.cos\Theta/2}}$

$\bold{\implies \frac{2sin^2\Theta/2 + 2sin\Theta/2.cos\Theta/2}{2cos^2\Theta/2 + 2sin\Theta/2.cos\Theta/2}}$

$\bold{\implies \frac{\cancel{2}sin\Theta/2\{\cancel{sin\Theta/2 + cos\Theta/2}\}}{\cancel{2}cos\Theta/2\{\cancel{cos\Theta/2 + sin\Theta/2}\}}}$

$\bold{\implies \frac{sin\Theta/2}{cos\Theta/2}}$

$\bold{\implies tan\Theta/2}$



I HOPE IT'S HELP YOU DEAR,
THANKS