Question

Value of $(1+\tan \theta +\sec \theta )(1+\cot \theta -co\sec \theta )$ is:

• A. $1$
• B. $-1$
• C. $2$
• D. $-4$

Answer 1

Answer: (C)

$2$

$(1+\tan \theta +\sec \theta )(1+\cot \theta -co\sec \theta )$
$(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta })(1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta })$
$\left(\frac{\cos \theta +\sin \theta +1}{\cos \theta }\right)\left(\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right)$
$=\frac{{(\cos \theta +\sin \theta )}^2-{\left(1\right)}^2}{\cos \theta \sin \theta }$
$=\frac{{\cos }^2\theta +{\sin }^2\theta +2\sin \theta \cos \theta -1}{\cos \theta \sin \theta }$
$=\frac{1+2\sin \theta \cos \theta -1}{\cos \theta \sin \theta }=\frac{2\sin \theta \cos \theta }{\cos \theta \sin \theta }=2$