# The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is

Question:

The value of $(1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)$ is

(a) 1
(b) 2
(c) 4
(d) 0

Solution:

The given expression is

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

Simplifying the given expression, we have

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

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

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

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

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

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

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

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

$=\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}$

$=2$

Therefore, the correct option is (b).