Question

Assertion (A): $\sin 1^0<\sin 1$
Reason (R): $f\left(x\right)=$ sinx is increasing function in $(0,\frac{\pi }{2})$

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true and R is not the correct explanation of A
• C. A is true and R is false
• D. A is false and R is true

Answer 1

The correct option is A Both A and R are true and R is the correct explanation of A

$1^0=\frac{\pi }{180}$ $1=\frac{180}{\pi }$
$1^0<1$
Therefore $\sin 1^0<\sin 1$ (as $\sin x$ is increasing in $(0,\frac{\pi }{2}))$