MCQ
${4 \over {1 + \sqrt 2 - \sqrt 3 }} = $
- ✓$2 + \sqrt 2 + \sqrt 6 $
- B$1 + \sqrt 2 + \sqrt 3 $
- C$3 + \sqrt 2 + \sqrt 3 $
- DNone of these
$ = {{4(1 + \sqrt 2 + \sqrt 3 )} \over {3 + 2\sqrt 2 - 3}} + {{\sqrt 6 (\sqrt 3 - \sqrt 2 )} \over {3 - 2}}$
$ = \sqrt 2 (1 + \sqrt 2 + \sqrt 3 ) = 2 + \sqrt 2 + \sqrt 6 $.
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$x \operatorname{cosec} \alpha-y \sec \alpha=\operatorname{kcot} 2 \alpha$ and $x \sin \alpha+y \cos \alpha=k \sin 2 \alpha$
respectively, then $\mathrm{k}^{2}$ is equal to :