MCQ 11 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: The value of $\sin(-690^\circ)\cos(-300^\circ)+\cos(-750^\circ)\sin(-240^\circ)=1.$
Reason: The values of $\sin$ and $\cos$ is negative in third and fourth quadrant respectively.
Assertion: The value of $\sin(-690^\circ)\cos(-300^\circ)+\cos(-750^\circ)\sin(-240^\circ)=1.$
Reason: The values of $\sin$ and $\cos$ is negative in third and fourth quadrant respectively.
- AAssertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- BAssertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- ✓Assertion is correct statement but Reason is wrong statement.
- DAssertion is wrong statement but Reason is correct statement.
Answer
View full question & answer→Correct option: C.
Assertion is correct statement but Reason is wrong statement.
$\sin(-690^\circ)=-\sin690^\circ=-\sin(2\times360^\circ-30^\circ)=-(-\sin30^\circ)=\frac{1}{2}$
$\cos(-300^\circ)=\cos300^\circ=\cos(360^\circ-60^\circ)=\cos60^\circ=\frac{1}{2}$
$\cos(-750^\circ)=\cos750^\circ=\cos(2\times360^\circ+30^\circ)=\cos30^\circ=\frac{\sqrt{3}}{2}$
$\sin(-240^\circ)=-\sin240^\circ=-\sin(180^\circ+60^\circ)$
$=(-\sin60^\circ)=\sin60^\circ=\frac{\sqrt{3}}{2}$
$\therefore\sin(-690^\circ)\cos(-300^\circ)+\cos(-750^\circ)\sin(-240^\circ)$
$=\Big(\frac{1}{2}\Big)\Big(\frac{1}{2}\Big)+\Big(\frac{\sqrt{3}}{2}\Big)\Big(\frac{\sqrt{3}}{2}\Big)$
$=\frac{1}{4}+\frac{3}{4}$
$=1.$
$\cos(-300^\circ)=\cos300^\circ=\cos(360^\circ-60^\circ)=\cos60^\circ=\frac{1}{2}$
$\cos(-750^\circ)=\cos750^\circ=\cos(2\times360^\circ+30^\circ)=\cos30^\circ=\frac{\sqrt{3}}{2}$
$\sin(-240^\circ)=-\sin240^\circ=-\sin(180^\circ+60^\circ)$
$=(-\sin60^\circ)=\sin60^\circ=\frac{\sqrt{3}}{2}$
$\therefore\sin(-690^\circ)\cos(-300^\circ)+\cos(-750^\circ)\sin(-240^\circ)$
$=\Big(\frac{1}{2}\Big)\Big(\frac{1}{2}\Big)+\Big(\frac{\sqrt{3}}{2}\Big)\Big(\frac{\sqrt{3}}{2}\Big)$
$=\frac{1}{4}+\frac{3}{4}$
$=1.$