Question

$(1+\tan \theta +\sec \theta )(1+\cot \theta -\csc \theta )=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

$(1+\cot \theta -\csc \theta )(1+\tan \theta +\sec \theta )$
$=(1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta })(1+\frac{\sin \theta }{\cos \theta }+\frac{2}{\cos \theta })$
$=\left(\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right)\left(\frac{\cos \theta +\sin \theta +1}{\cos \theta }\right)$
$=\frac{\sin \theta \cos \theta +{\sin }^2\theta +\sin \theta +{\cos }^2\theta +\sin \theta \cos \theta +\cos \theta -\cos \theta -\sin \theta -1}{\sin \theta \cos \theta }$
$=\frac{2\sin \theta \cos \theta +{\sin }^2\theta +{\cos }^2\theta -1}{\sin \theta \cos \theta }$
We know that $[\because {\sin }^2\theta +{\cos }^2\theta =1]$
$=\frac{2\sin \theta \cos \theta +1-1}{\sin \theta \cos \theta }$
$=\frac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }$
$=2$