MCQ
The value of $(1+\cot\theta-\text{cosec }\theta)(1+\tan\theta+\sec\theta)$ is
- A$1$
- ✓$2$
- C$4$
- D$0$
✓
Answer
Correct option: B.
$2$
$(1+\cot\theta-\text{cosec }\theta)(1+\tan\theta+\sec\theta)$
$=\Big(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\Big)\Big(1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}\Big)$
$=\frac{(\sin\theta+\cos\theta-1)(\cos\theta+\sin\theta+1)}{\sin\theta\times\cos\theta}$
$=\frac{\{(\sin\theta+\cos\theta)-1\}\{(\cos\theta+\sin\theta)+1\}}{\sin\theta\cos\theta}$
$=\frac{(\cos\theta+\sin\theta)^2-1}{\sin\theta\cos\theta}$
$=\frac{\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}$
$=\frac{1+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}=\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}$
$=2$
$=\Big(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\Big)\Big(1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}\Big)$
$=\frac{(\sin\theta+\cos\theta-1)(\cos\theta+\sin\theta+1)}{\sin\theta\times\cos\theta}$
$=\frac{\{(\sin\theta+\cos\theta)-1\}\{(\cos\theta+\sin\theta)+1\}}{\sin\theta\cos\theta}$
$=\frac{(\cos\theta+\sin\theta)^2-1}{\sin\theta\cos\theta}$
$=\frac{\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}$
$=\frac{1+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}=\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}$
$=2$
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