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

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