MCQ
Choose the correct option. Justify your choice : $(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) $
$=\Big(1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}\Big)\Big(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\Big) $
$=\Big(\frac{\cos\theta+\sin\theta+1}{\cos\theta}\Big)\Big(\frac{\sin\theta+\cos\theta-1}{\sin\theta}\Big)$
$=\frac{(\cos\theta+\sin\theta)^2-(1)^2}{\cos\theta.\sin\theta} $
$=\frac{\cos^2\theta+\sin^2+2\cos\theta\sin\theta-1}{\cos\theta.\sin\theta} $
$=\frac{1+2\cos\theta\sin\theta-1}{\cos\theta.\sin\theta} $
$[\because\sin^2\theta+\cos^2\theta=1] $
$=\frac{2\cos\theta\sin\theta}{\cos\theta.\sin\theta}=2 $
$=\Big(1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}\Big)\Big(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\Big) $
$=\Big(\frac{\cos\theta+\sin\theta+1}{\cos\theta}\Big)\Big(\frac{\sin\theta+\cos\theta-1}{\sin\theta}\Big)$
$=\frac{(\cos\theta+\sin\theta)^2-(1)^2}{\cos\theta.\sin\theta} $
$=\frac{\cos^2\theta+\sin^2+2\cos\theta\sin\theta-1}{\cos\theta.\sin\theta} $
$=\frac{1+2\cos\theta\sin\theta-1}{\cos\theta.\sin\theta} $
$[\because\sin^2\theta+\cos^2\theta=1] $
$=\frac{2\cos\theta\sin\theta}{\cos\theta.\sin\theta}=2 $
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