Question

What is $\frac{(1+\sec \theta -\tan \theta )\cos \theta }{(1+\sec \theta +\tan \theta )(1-\sin \theta )}$ equal to

• A. $1$
• B. $2$
• C. $\tan \theta$
• D. $\cot \theta$

Answer 1

The correct option is A $1$

We have,

$=\frac{(1+\sec \theta -\tan \theta )\cos \theta }{(1+\sec \theta +\tan \theta )(1-\sin \theta )}$

$=\frac{(1+\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta })\cos \theta }{(1+\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta })(1-\sin \theta )}$

$=\frac{(\cos \theta +1-\sin \theta )\cos \theta }{(\cos \theta +1+\sin \theta )(1-\sin \theta )}$

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

$=\frac{({\cos }^2\theta +\cos \theta -\sin \theta \cos \theta )}{(\cos \theta -\cos \theta \sin \theta +{\cos }^2\theta )}[\because {\cos }^2\theta =1-{\sin }^2\theta ]$

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

$=1$

Hence, this is the answer.