MCQ
If $\tan \frac{\theta }{2} = t,$then $\frac{{1 - {t^2}}}{{1 + {t^2}}}$is equal to
- ✓$\cos \theta $
- B$\sin \theta$
- C$\sec \theta $
- D$\cos 2\theta $
$ \Rightarrow \frac{{1 - {{\tan }^2}\frac{\theta }{2}}}{{1 + {{\tan }^2}\frac{\theta }{2}}} = \cos \theta $.
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