MCQ
$\cos ({\tan ^{ - 1}}x) = $
- A$\sqrt {1 + {x^2}} $
- ✓$\frac{1}{{\sqrt {1 + {x^2}} }}$
- C$1 + {x^2}$
- Dએકપણ નહીં.
$\therefore \,\,\cos \theta = \frac{1}{{\sqrt {1 + {{\tan }^2}\theta } }} = \frac{1}{{\sqrt {1 + {x^2}} }}$
Hence $\cos \theta = \cos \,({\tan ^{ - 1}}x) = \frac{1}{{\sqrt {1 + {x^2}} }}$.
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