Question

If $p = \tan 1^\circ$ , $q = \tan 1$ (in radians), then which of the following is true?
A. $p = q$
B. $p < q$
C. $p > q$
D. $2p = 3q$

Answer 1

In the above given question, we are given two equations $p$ and $q$ which are both functions of the trigonometric function called tangent. The difference between these two given functions is that the first function, the angle is in the units of degree where as the second function has the unit of the angle in radians. We have to determine the correct relation between these two functions from the four given options.

Complete step by step answer:
We are given that, two tangential functions $p$ and $q$ such as,
$\Rightarrow p = \tan 1^\circ$
And
$\Rightarrow q = \tan 1$ (in radian units)
We have to determine the correct relation between the functions $p$ and $q$. Since we know that $180^\circ = \pi$ radians, therefore we can write
$\Rightarrow \dfrac{{180^\circ }}{\pi } = 1$ radian
Substituting $\pi = 3.14$ and calculating, we get the above equation as,
$\Rightarrow 57.32^\circ = 1$ radian
Therefore now we have the two functions as $p = \tan 1^\circ$ and $q = \tan 57.32^\circ$. Since we know that the tangent function is increasing in the interval $\left[ {0,\dfrac{\pi }{2}} \right]$.

Therefore, for the interval $\left[ {0,\dfrac{\pi }{2}} \right]$ , that gives us
$\Rightarrow {x_1} > {x_2} \Leftrightarrow \tan {x_1} > \tan {x_2}$
Now since we have,
$\Rightarrow 57.32^\circ > 1^\circ$
Hence, from the above statement, that gives us
$\Rightarrow \tan 57.32^\circ > \tan 1^\circ$
Which is actually the substitution of
$\Rightarrow \tan 1 > \tan 1^\circ$
Therefore, we can write the above relation as,
$\Rightarrow q > p$
We can also write it as,
$\therefore p < q$
That is the required relation between $p$ and $q$ .

Therefore, the correct option is B.

Note: The tangent function of an angle is defined as the trigonometric ratio between the adjacent i.e. the perpendicular side and the opposite side i.e. the base of a right angled triangle containing that angle.It is also calculated as the ratio of two other trigonometric functions, the sine and cosine functions as $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.