Question

Choose the correct answer and justify.
$(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\theta )=$

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

Answer: (C)

$2$

$(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\theta )$
$=\{1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }\}\times \{1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }\}$
$=\left\{\frac{\cos \theta +\sin \theta +1}{\cos \theta }\right\}\times \left\{\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right\}$
$=\frac{\{\cos \theta +\sin \theta +1\}\times \{\sin \theta +\cos \theta -1\}}{\cos \theta \times \sin \theta }$
$=\frac{(\cos \theta +\sin \theta {)}^2-(1{)}^2}{\cos \theta \times \sin \theta }[\because (a+b\left)\right(a-b)=a^2-b^2]$
$=\frac{{\cos }^{\theta }+{\sin }^2\theta +2\cos \theta \sin \theta -1}{\cos \theta \times \sin \theta }$
$=\frac{1+2\cos \theta \sin \theta -1}{\cos \theta \times \sin \theta }=2$