Question

$\mathrm{if}\cos \theta =\frac{12}{13}\mathrm{and}\theta \in \left(0,\frac{\pi }{2}\right],\mathrm{then}144\left(\tan (-\theta )\times \sec (-\theta )\right]$

Answer 1

$\mathrm{Given},\theta \in \left(0,\frac{\pi }{2}\right]$
This means $\theta$ lies in first quadrant.
$\cos \theta =\frac{12}{13}\Rightarrow \sin \theta =\sqrt{1-\frac{{12}^2}{{13}^2}}\Rightarrow \sin \theta =\frac{5}{13}\Rightarrow \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{5}{12}\mathrm{and}\sec \theta =\frac{1}{\cos \theta }=\frac{13}{12}$
Also we know,
$\tan (-\theta )=-\tan \theta \mathrm{and}\sec (-\theta )=\sec \theta$
Thus, $\tan (-\theta )\times \sec (-\theta )\Rightarrow -\tan \theta \times \sec \theta$
Substituting values we get,
$-\tan \theta \times \sec \theta =-\frac{5}{12}\times \frac{13}{12}=-\frac{65}{144}\Rightarrow 144\tan \theta \times \sec \theta =\frac{-65}{144}\times 144\Rightarrow 144\tan \theta \times \sec \theta =-65$

​Hence the correct answer is Option B.