Question
State True or False for the following statement:
$\sin10^\circ$ is greater than $\cos10^\circ.$
$\sin10^\circ$ is greater than $\cos10^\circ.$
✓
Answer
False.
Solution:
$$If $\sin10^\circ>\cos10^\circ$
$\Rightarrow\sin10^\circ>\cos(90^\circ-80^\circ)$
$\Rightarrow\sin10^\circ>\sin80^\circ$ Which is not possible, because value of sine is in increasing order .
Hence, the statement is 'False'.
Solution:
$$If $\sin10^\circ>\cos10^\circ$
$\Rightarrow\sin10^\circ>\cos(90^\circ-80^\circ)$
$\Rightarrow\sin10^\circ>\sin80^\circ$ Which is not possible, because value of sine is in increasing order .
Hence, the statement is 'False'.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $n(A)=p$ and $n(B)=q$ then $n(A \times B )=2^{p q}$
→Let A = {{1, 2,3}, {4, 5}, {6, 7, 8}}. Then the following is true or false:
$\phi\in\text{A}.$
→$\phi\in\text{A}.$
$\lim _{x \rightarrow 1} \frac{x^2-\sqrt{x}}{\sqrt{x}-1}=3$
→State Whether the statements are True or False. Justify.
The shortest distance from the point $(2, -7)$ to the circle $x^2 + y^2 - 14x - 10y - 151 = 0$ is equal to $5.$
[Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point]
→The shortest distance from the point $(2, -7)$ to the circle $x^2 + y^2 - 14x - 10y - 151 = 0$ is equal to $5.$
[Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point]
If $f: X \rightarrow R , f(x)=x^3+1$ then $f(9)=730$.
→If $z$ is a complex number such that $z \neq 0$ and Re $(z) = 0,$ then $Im (z^2) = 0.$
→On throwing two dice, the probability of obtaining a number greater than 3 on at least one dice is $\frac{3}{4}.$
→State which of the following statements is True or False.
Solution set of $\text{x}\geq0$ and $\text{y}\leq0.$
→Solution set of $\text{x}\geq0$ and $\text{y}\leq0.$
The probability of intersection of two events A and B is always less than or equal to those favourable to the event A.
Let A = {a, b, {c, d}, e}. Then the following statements is false and why?
$\{\text{a, b, e}\}\in\text{A}$
→$\{\text{a, b, e}\}\in\text{A}$