Question

State whether the following statement is true/false

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

• A. True
• B. False

Answer 1

The correct option is A True

Accordin to question,

$$\begin{array}{c}Wehave,.........\frac{1+sec\theta -tan\theta }{1+sec\theta -tan\theta }[weknow,{\sec }^2\theta -{\tan }^2\theta =1\\ \Rightarrow \frac{({\sec }^2\theta -{\tan }^2\theta )+sec\theta -tan\theta }{1+\sec \theta +\tan \theta }\\ \Rightarrow \frac{(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )+(sec\theta -tan\theta )}{1+\sec \theta +\tan \theta }\\ \Rightarrow \frac{(\sec \theta -\tan \theta )+(\sec \theta +\tan \theta +1)}{1+\sec \theta +\tan \theta }\\ \Rightarrow (\sec \theta -\tan \theta )=\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }\\ \therefore \frac{1-\sin \theta }{\cos \theta }prove\\ Sowecansaythatquestionistrue.andcorrectoptionisA.\end{array}$$