Question

Which of the following options is equal to $\frac{\sin \theta +1-\cos \theta }{\cos \theta -1+\sin \theta }?$

• A. $\frac{\cos \theta }{1-\sin \theta }$
• B. $\frac{1+\sin \theta }{\cos \theta }$
• C. $\frac{1+\sin \theta }{1-\sin \theta }$
• D. $\tan \theta +\sec \theta$

Answer 1

Answer: (A), (B), (D)

The correct options are
A

$\frac{\cos \theta }{1-\sin \theta }$

B

$\frac{1+\sin \theta }{\cos \theta }$

D

$\tan \theta +\sec \theta$

$\frac{sin\theta +1-cos\theta }{cos\theta -1+sin\theta }$

Dividing the numerator and denominator by $cos\theta$ we get,

$\frac{\frac{sin\theta }{cos\theta }+\frac{1}{cos\theta }-\frac{cos\theta }{cos\theta }}{\frac{cos\theta }{cos\theta }-\frac{1}{cos\theta }+\frac{sin\theta }{cos\theta }}$

$=\frac{tan\theta +sec\theta -1}{1-sec\theta +tan\theta }$

$\Rightarrow \frac{(tan\theta +sec\theta )-(se{c}^2\theta -ta{n}^2\theta )}{1-sec\theta +tan\theta }....(1+{\tan }^2\theta ={\sec }^2\theta )$

$=\frac{(\tan \theta +\sec \theta )-(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )}{1-\sec \theta +\tan \theta }$

$=\frac{(\tan \theta +\sec \theta )(1-\sec \theta +\tan \theta )}{(1-\sec \theta +\tan \theta )}$

$=\tan \theta +\sec \theta$

$=\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }$

$=\frac{\sin \theta +1}{\cos \theta }$

[Multiplying the numerator and denominator by $(1-\sin \theta )$]
$=\frac{1+\sin \theta }{\cos \theta }\times \frac{1-\sin \theta }{1-\sin \theta }$

$=\frac{1-{\sin }^2\theta }{\cos \theta (1-\sin \theta )}$

$=\frac{{\cos }^2\theta }{\cos \theta (1-\sin \theta )}$

$=\frac{\cos \theta }{1-\sin \theta }$