MCQ
$(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)=$
- A$0$
- B$1$
- ✓$2$
- D$-1$
✓
Answer
Correct option: C.
$2$
$(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)$
$=1+\cot\theta-\text{cosec }\theta+\tan\theta+\cot\theta\tan\theta$
$=-\tan\theta\text{ cosec }\theta+\sec\theta+\sec\theta\cot\theta-\sec\theta\text{ cosec }\theta$
$=1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}+\frac{\sin\theta}{\cos\theta}+1-\frac{\sin\theta}{\cos\theta}\times\frac{1}{\sin\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\cos\theta}\times\frac{\cos\theta}{\sin\theta}-\frac{1}{\cos\theta}\times\frac{1}{\sin\theta}$
$=2+\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}-\frac{1}{\sin\theta}-\frac{1}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{1}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}=2$
$=1+\cot\theta-\text{cosec }\theta+\tan\theta+\cot\theta\tan\theta$
$=-\tan\theta\text{ cosec }\theta+\sec\theta+\sec\theta\cot\theta-\sec\theta\text{ cosec }\theta$
$=1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}+\frac{\sin\theta}{\cos\theta}+1-\frac{\sin\theta}{\cos\theta}\times\frac{1}{\sin\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\cos\theta}\times\frac{\cos\theta}{\sin\theta}-\frac{1}{\cos\theta}\times\frac{1}{\sin\theta}$
$=2+\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}-\frac{1}{\sin\theta}-\frac{1}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{1}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}=2$
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