(1 + tan θ + sec θ) (1 + cot θ − cosec θ) =


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

(a) 0
(b) 1
(c) 1
(d) −1


The given expression is $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$.

Simplifying the given expression, we have

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

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

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

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

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

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

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

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

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

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


Disclaimer: None of the given options match with the answer.




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