If $\;\frac{{3\,\pi }}{4} < \alpha < \pi ,\,{\rm{\;then}}\;\,\sqrt {{\rm{cose}}{{\rm{c}}^2}\,\alpha + 2\,\,\cot \,\,\alpha } {\rm{\;}}$ is equal to

$ - 1 + {\rm{cot}}\;\,\,\alpha $

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

\;\frac{{3\,\pi }}{4} < \alpha < \pi ,\,\mathrm{\;then}\;\,\sqrt {\mathrm{cose}{\mathrm{c}^2}\,\alpha + 2\,\,\cot \,\,\alpha } \mathrm{\;}

is equal to

View answer and explanation

Correct option: C

- 1 - \cot \,\,\alpha

Option C (

- 1 - \cot \,\alpha

) is correct because it considers the range of

\alpha

and simplifies the expression accordingly.