Question

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

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

Answer 1

The correct option is C $\frac{1+sin\theta }{cos\theta }$

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