Question
यदि $\frac{{3\pi }}{4} < \alpha < \pi ,$ हो, तब $\sqrt {{\rm{cose}}{{\rm{c}}^2}\alpha + 2\cot \alpha } $ बराबर है
$= \sqrt {1 + {{\cot }^2}\alpha + 2\cot \alpha } = \,\,|1 + \cot \alpha |$
लेकिन $\frac{{3\pi }}{4} < \alpha < \pi \Rightarrow \cot \alpha < - 1$
$\Rightarrow 1 + \cot \alpha < 0$
अतः $|1 + \cot \alpha | = - (1 + \cot \alpha )$.
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कथन$-2:$ किसी प्राकृत संख्या $n$ के लिए $\mathop \sum \limits_{k = 1}^n \left( {{k^3} - {{\left( {k - 1} \right)}^3}} \right) = {n^3}$ है।