MCQ
$(\sec \theta-\tan \theta)^2=$ ?
- ✓$\frac{1-\sin \theta}{1+\sin \theta}$
- B$\frac{1+\sin \theta}{1-\sin \theta}$
- C$\frac{1+\cos \theta}{1-\cos \theta}$
- D$\frac{1-\cos \theta}{1+\cos \theta}$
✓
Answer
Correct option: A.
$\frac{1-\sin \theta}{1+\sin \theta}$
$(\sec \theta-\tan \theta)^2$
$=\left(\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}\right)^2$
$=\left(\frac{1-\sin \theta}{\cos \theta}\right)^2=\frac{(1-\sin \theta)^2}{\cos ^2 \theta}$
$=\frac{(1-\sin \theta)^2}{\left(1-\sin ^2 \theta\right)}$
$=\frac{(1-\sin \theta)^2}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{1-\sin \theta}{1+\sin \theta}$
$=\left(\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}\right)^2$
$=\left(\frac{1-\sin \theta}{\cos \theta}\right)^2=\frac{(1-\sin \theta)^2}{\cos ^2 \theta}$
$=\frac{(1-\sin \theta)^2}{\left(1-\sin ^2 \theta\right)}$
$=\frac{(1-\sin \theta)^2}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{1-\sin \theta}{1+\sin \theta}$
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