Question

If ${180}^o<\theta <{270}^{o}andsin\theta =-\frac{5}{13},$ then $5{\cot }^2\theta +12tan\theta +13cosec\theta =$

• A. $\frac{5}{144}$
• B. $\frac{144}{5}$
• C. $-\frac{5}{144}$
  
• D. $-\frac{144}{5}$
  

Answer 1

The correct option is B $\frac{144}{5}$

$\cos \theta =\pm \sqrt{1-{\sin }^2\theta }=-\sqrt{1-\frac{25}{169}}=-\sqrt{\frac{144}{169}}=\frac{-12}{13}(\theta =II{I}^{rd}quad\cos \theta <9)$

$\Rightarrow \cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{12}{4}$,

$\tan \theta =\frac{1}{\cot \theta }=\frac{5}{12}$

$\cot \theta =\frac{1}{\sin \theta }=\frac{-13}{5}$

Expression$=5\left(\frac{144}{25}\right)+12\left(\frac{5}{12}\right)-\frac{13\times 5}{13}=\frac{144}{5}$.