Question

If $1+\tan 18°\tan 9°=\sec \theta (0<\theta <90°)$, then which of the following is/are true?

• A. $3\theta$ lies in $3^{\text{rd}}$ quadrant
• B. $\sin 5\theta +\cos 5\theta =1$
• C. $\frac{1+\tan \theta /2}{1-\tan \theta /2}=\cot 2\theta$
• D. $\sec 10\theta +1=0$

Answer 1

Answer: (B), (C), (D)

$\sec 10\theta +1=0$

$1+\tan 18°\tan 9°=1+\frac{\sin 18°}{\cos 18°}\frac{\sin 9°}{\cos 9°}=\frac{\cos 18°\cos 9°+\sin 18°\sin 9°}{\cos 18°\cos 9°}=\frac{\cos (18°-9°)}{\cos 18°\cos 9°}=\frac{1}{\cos 18°}=\sec 18°\therefore \theta =18°$

$\left(a\right)$ $3\theta =54°$ lies in $1^{\text{st}}$ quadrant

$\left(b\right)$ $\sin \theta +\cos 5\theta =\sin 90°+\cos 90°=1$

$\left(c\right)$ $\frac{1+\tan 9°}{1-\tan 9°}=\tan (45°+9°)=\tan 54°=\cot 36°$

$\left(d\right)$ $\sec 10\theta +1=\sec 180°+1=-1+1=0$