Question

$|\tan \theta +\sec \theta |=|\tan \theta |+|\sec \theta |,0\le \theta \le 2\pi$ is possible only if-

• A. $\theta ϵ[0,\pi ]-\left\{\frac{\pi }{2}\right\}$
• B. $\theta ϵ[0,\pi ]$
• C. $\theta ϵ[0,\frac{\pi }{2})$
• D. $(0,\frac{\pi }{2}]$

Answer 1

The correct option is A $\theta ϵ[0,\pi ]-\left\{\frac{\pi }{2}\right\}$

$\mid \tan \theta +\sec \theta \mid =\mid \tan \theta \mid +\sec \theta$

Case 1:-

If $\theta \in [0,\frac{\pi }{2})$ then $\tan \theta =+ve,\sec \theta =+ve$

$\tan \theta +\sec \theta =+ve$

$\therefore \mid \sec \theta +\tan \theta \mid =\mid \tan \theta \mid +\mid \sec \theta \mid$

Case 2:-

$\theta \in (\frac{\pi }{2},\pi ]$

$\tan \theta =-ve,\sec \theta =-ve$

$\mid \sec \theta +\tan \theta \mid =+ve$

$\mid \sec \theta \mid =+ve\mid \tan \theta \mid =+ve$

$\mid \sec \theta +\tan \theta \mid =\mid \sec \theta \mid +\mid \tan \theta \mid$

Case 3:-

$\theta \in [\pi ,\frac{3\pi }{2}]$

$\tan \theta =+ve,\sec \theta =-ve$

$\mid \sec \theta +\tan \theta \mid =\tan \theta -\sec \theta$

$\mid \tan \theta \mid +\mid \sec \theta \mid =\sec \theta +\tan \theta$

$\mid \sec \theta +\tan \theta \mid \ne \mid \sec \theta \mid +\mid \tan \theta \mid$

Case 4:-

$\theta \in (\frac{3\pi }{2},2\pi ]$

$\tan \theta =-ve,\sec \theta =+ve$

$\mid \sec \theta +\tan \theta \mid =\sec \theta -\tan \theta$

$\mid \sec \theta \mid =\sec \theta \mid \tan \theta \mid =\tan \theta$

$\mid \sec \theta +\tan \theta \mid \ne \mid \sec \theta \mid +\mid \tan \theta \mid$

$\theta \ne \frac{\pi }{2}$ because in that case $\tan \theta =\infty$