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8. Introduction to Trigonometry
hard
$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=.......$
A
$0$
B
$2$
C
$1$
D
$-1$
Solution
$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$
$=\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right)$
$=\left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right)\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right)$
$=\frac{(\sin \theta+\cos \theta)^{2}-(1)^{2}}{\sin \theta \cos \theta}$
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}$
$=\frac{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}$
$=\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}=2$
Standard 10
Mathematics
