Question

i. $\tan {720}^{\circ }-\cos {270}^{\circ }-\sin {150}^{\circ }\cos {120}^{\circ }=\frac{1}{4}$
ii. $\sin {780}^{\circ }\sin {480}^{\circ }+\cos {120}^{\circ }\sin {150}^{\circ }=\frac{1}{2}$

which of the above statements are true.

• A. only $1st$ statement is true
• B. only $2nd$ statement is true
• C. both $1st$ and $2nd$ statements are true
• D. none of them are true

Answer 1

The correct option is C both $1st$ and $2nd$ statements are true

(i) $\tan {720}^{\circ }-\cos {270}^{\circ }-\sin {150}^{\circ }\cos {120}^{\circ }$

$=\tan ({360}^{\circ }\times 2)-\cos ({360}^{\circ }-{90}^{\circ })-\sin ({180}^{\circ }-{30}^{\circ })\cos ({90}^{\circ }+{30}^{\circ })$

$=0-\cos \left({90}^{\circ }\right)+\sin {30}^{\circ }\sin {30}^{\circ }=-0+\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

(ii) $\sin {780}^{\circ }\sin {480}^{\circ }+\cos {120}^{\circ }\sin {150}^{\circ }$

$=\sin (2\times {360}^{\circ }+{60}^{\circ })\sin ({360}^{\circ }+{120}^{\circ })+\cos ({180}^{\circ }-{60}^{\circ })\sin ({180}^{\circ }-{30}^{\circ })$

$\Rightarrow \sin {60}^{\circ }\sin \left({120}^{\circ }\right)-\cos {60}^{\circ }\sin {30}^{\circ }\Rightarrow \sin {60}^{\circ }\sin ({90}^{\circ }+{30}^{\circ })-\cos {60}^{\circ }\sin {30}^{\circ }$

$=\sin {60}^{\circ }\cos {30}^{\circ }-\cos {60}^{\circ }\sin {30}^{\circ }$

$\Rightarrow \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$