Question

$(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\theta )$ is equal to

• A. $0$
• B. $2$
• C. $1$
• D. $-1$

Answer 1

Answer: (B)

$2$

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