Question

If $\cos \alpha =\frac{12}{13},\cot \beta =\frac{24}{7},{90}^{\circ }<\alpha <{180}^{\circ }$ and ${180}^{\circ }<\beta <{270}^{\circ }$ then the quadrant in which $\alpha +\beta$ lies

• A. $Q_1$
• B. $Q_2$
• C. $Q_3$
• D. $Q_4$

Answer 1

The correct option is D $Q_4$

Given $cos\alpha =\frac{12}{13},cot\beta =\frac{24}{7}$

$[{90}^{\circ }<\alpha <{180}^{\circ }cos(x)has-vevalues]$

$sin\alpha =\sqrt{1-(\frac{12}{13}{)}^2}=\frac{5}{13}$

$cosec\beta =\sqrt{1+(\frac{24}{7}{)}^2}=\frac{-25}{7}$

$[si{n}^2\alpha +cos{\alpha }^2-11+co{t}^2\beta =cose{c}^2\beta ]$

$\Rightarrow sin\beta =\frac{-7}{25},cos\beta =\frac{-24}{25}$

Now, $sin(\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta$

$=\left(\frac{5}{13}\right)\left(\frac{-24}{25}\right)+\left(\frac{-12}{13}\right)\left(\frac{-7}{25}\right)<0$

$\Rightarrow (\alpha +\beta )$ lies in quadrant IV or III ...(i)

$cos(\alpha +\beta )=cos\alpha cos\beta -sin\alpha sin\beta$

$=\left(\frac{-12}{13}\right)\left(\frac{-24}{25}\right)-\left(\frac{5}{13}\right)\left(\frac{-7}{25}\right)>0$

$\Rightarrow (\alpha +\beta )$ lies in Quadrant I or IV ...(ii)

From (1), (2) $(\alpha +\beta )$ lie is quadrant IV

$\therefore$ option D is correct.