Question

If $3\pi /4<\alpha <\pi$, then $\sqrt{(\cos ec2\alpha +2cot\alpha )}$is equal to

• A. $1+cot\alpha$
• B. $1–cot\alpha$
• C. $-1–cot\alpha$
• D. $-1+cot\alpha$

Answer 1

The correct option is C

$-1–cot\alpha$

To find the value for $\sqrt{\mathbf{(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{e}{\mathbf{c}}^{\mathbf{2}}\mathbf{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{\alpha }\mathbf{)}\mathbf{:}}$

Using the identity

$\begin{array}{rcl}\cos e{c}^{2}A–co{t}^{2}A& =& 1\\ \cos e{c}^{2}A& =& co{t}^{2}A+1\\ & & \\ & & \end{array}$

$\begin{array}{rcl}\surd (\cos e{c}^2\alpha +2cot\alpha )& =& \surd [1+co{t}^2\alpha +2cot\alpha ]\\ & =& \surd {(1+cot\alpha )}^2\\ & =& |1+cot\alpha |\end{array}$

Given that $3\frac{\pi }{4}<\alpha <\pi .$

In this interval $cot\alpha$is negative.

Therefore, $|1+cot\alpha |=-1–cot\alpha .$option (C) is correct.