Question

Choose the correct option:

I: $\cos \theta \cdot \cot ({300}^{\circ }-\theta )\cdot \cos ({300}^{\circ }+\theta )=\frac{1}{4}\cos 3\theta$

II: $\tan \theta ({300}^{\circ }-\theta )\cdot \tan ({300}^{\circ }+\theta )=\frac{1}{4}\tan 3\theta$

• A. only I is true
• B. only II is true
• C. Both I and II are true
• D. Neither I nor II are true

Answer 1

The correct option is A only I is true

I: $\cos \theta \cdot \cos (300-\theta )\cdot \cos (300+\theta )$

$=\cos \theta \cdot \cos (360-(60+\theta \left)\right)\cdot \cos (360-(60-\theta \left)\right)$

$=cos\theta \cdot \cos (60+\theta )\cdot \cos (60-\theta )$

$=\frac{1}{2}\cos \theta (\cos 120+\cos 20)$

$=\frac{1}{2}\cos \theta \left(\cos \right(90+30)+2{\cos }^2\theta -1)$

$=\frac{1}{2}\cos \theta (-\sin 30+2{\cos }^2\theta -1)$

$=\frac{1}{2}\cos \theta (-\frac{1}{2}+2{\cos }^2\theta -1)$

$=\frac{1}{2}\cos \theta (-\frac{3}{2}+2{\cos }^2\theta )$

$=\frac{1}{4}(-3\cos \theta +4{\cos }^3\theta )=\frac{1}{4}\cos 3\theta$

II: $\sin \theta \cdot \sin (300-\theta )\cdot \sin (300+\theta )$

$=\sin \theta \cdot \sin (360-(60+\theta \left)\right)\cdot \sin (360-(60-\theta \left)\right)$

$=\sin \theta \cdot \sin (60+\theta )\cdot \sin (60-\theta )$

$=\frac{1}{2}\sin \theta (\cos 2\theta -\cos 120)$

$=\frac{1}{2}\sin \theta (1-2{\sin }^2\theta -\cos (90+30\left)\right)$

$=\frac{1}{2}\sin \theta (1-2{\sin }^2\theta +\sin 30)$

$=\frac{1}{2}\sin \theta (\frac{3}{2}-2{\sin }^2\theta )$

$=\frac{1}{4}(3\sin \theta -4{\sin }^3\theta )=\frac{1}{4}\sin 3\theta$

Therefore ,$\tan \theta \cdot \tan (300-\theta )\cdot \tan (300+\theta )$

$=\frac{\sin \theta \cdot \sin (300-\theta )\cdot \sin (300+\theta )}{\cos \theta \cdot \cos (300-\theta )\cdot \cos (300+\theta )}$

$=\frac{\frac{1}{4}\sin 3\theta }{\frac{1}{4}\cos 3\theta }=\tan 3\theta$