# (1 + tan θ + sec θ) (1 + cot θ – cosec θ) = ?

Question:

(1 + tan θ + sec θ) (1 + cot θ – cosec θ) = ?
(a) –1
(b) 0
(c) 1
(d) 2

Solution:

$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$

$=1+\cot \theta-\operatorname{cosec} \theta+\tan \theta+\tan \theta \cot \theta-\tan \theta \operatorname{cosec} \theta+\sec \theta+\sec \theta \cot \theta-\sec \theta \operatorname{cosec} \theta$

$=1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta}+\tan \theta \times \frac{1}{\tan \theta}-\frac{\sin \theta}{\cos \theta} \times \frac{1}{\sin \theta}+\frac{1}{\cos \theta}+\frac{1}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}-\frac{1}{\cos \theta \sin \theta}$

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

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

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

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

$=2$

Hence, the correct answer is option (d).