Question

$\frac{2\sec \theta +3\tan \theta +5\sin \theta -7\cos \theta +5}{2\tan \theta +3\sec \theta +5\cos \theta -7\sin \theta +8}=$

• A. $\tan \frac{\theta }{2}$
• B. $\cot \frac{\theta }{2}$
• C. $\sec \frac{\theta }{2}$
• D. $co\sec \frac{\theta }{2}$

Answer 1

Answer: (A)

$\tan \frac{\theta }{2}$

Let $c=\cos \theta ,s=\sin \theta$
$\therefore E=\frac{2+3s+5cs-7s^2+5c}{2s+3+5c^2+7cs+8c}=\frac{s(5c+3)-(7c+2)(c-1)}{s(7c+2)+(c+1)(5c+3)}$
$s^2=1-c^2=(1+c)(1-c)$
$\therefore \frac{s}{1-c}=\frac{1+c}{s}=\lambda$
$\therefore E=\frac{(1-c)\lambda (5c+3)+(1-c)(7c+2)}{s(7c+2)+\lambda s(5c+3)}=\frac{1-c}{s}.\frac{(7c+2)+\lambda (5c+3)}{7c+2)+\lambda (5c+3)}=\frac{1-c}{s}=\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=\tan \frac{\theta }{2}$