If $\;0^\circ < {\rm{\theta }} < 180^\circ ,\;\;{\rm{then}}\,\,\sqrt {2\,\, + \,\,\sqrt {2\,\, + \,\,\sqrt {2\,\, + \,\,... + \sqrt {2\,\,\left( {1 + \cos \,\,{\rm{\theta }}} \right),} } } } $ then being n number of ${\rm{2'}}\,{\rm{s}}$, is equal to

$2\,\,\cos \,\,\left( {\frac{{\rm{\theta }}}{{{2^n}}}} \right)$

$2\,\,\cos \,\,\left( {\frac{{\rm{\theta }}}{{{2^{n - 1}}}}} \right)$

$2\,\,\cos \,\,\left( {\frac{{\rm{\theta }}}{{{2^{n + 1}}}}} \right)$

JEE MATHEMATICS MCQ: MULTIPLE AND SUBMULTIPLE ANGLES

1 Medium

1 of 1

Application +4 | -1

\;0^\circ < \mathrm{\theta } < 180^\circ ,\;\;\mathrm{then}\,\,\sqrt {2\,\, + \,\,\sqrt {2\,\, + \,\,\sqrt {2\,\, + \,\,... + \sqrt {2\,\,\left( {1 + \cos \,\,\mathrm{\theta }} \right),} } } }

then being n number of

\mathrm{2'}\,\mathrm{s}

, is equal to

View answer and explanation

Correct option: A

2\,\,\cos \,\,\left( {\frac{\mathrm{\theta }}{{{2^n}}}} \right)

Option A (

2\cos \left( {\frac{\mathrm{\theta }}{{{{2^n}}}}} \right)

) is correct because the nested square roots simplify using trigonometric identities to this expression.