Question

$\frac{\sec \theta +\tan \theta -1}{\tan \theta -\sec \theta +1}=$

• A. $\sec \theta -\tan \theta$
• B. $\tan \theta -\sec \theta$
• C. $\sec \theta +\tan \theta$
• D. $1$

Answer 1

Answer: (C)

$\sec \theta +\tan \theta$

$\begin{array}{c}\frac{\sec \theta +\tan \theta -1}{\tan \theta -\sec \theta +1}\\ now,\\ =\frac{1+\sin \theta -\cos \theta }{\sin \theta +\cos \theta -1}\\ =\frac{1+2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}-1+2\sin \frac{\theta }{2}}{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}+1-2{\sin }^2\frac{\theta }{2}-1}\\ =\frac{2\sin \frac{\theta }{2}\left(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\right)}{2\sin \frac{\theta }{2}\left(\cos \frac{\theta }{2}-\sin \frac{\theta }{2}\right)}\\ =\left(\sec \theta +\tan \theta \right)\\ OptionCiscorrectanswer.\end{array}$