Question

How do you evaluate $\csc {75^0}$ ?

Answer 1

Hint : Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. Thus the given function can be converted in the form of sine easily. First we find the value of sine function then taking the reciprocal of sine we get the cosecant value. Using this information, we can find the solution of $\csc {75^0}$ .

Complete step-by-step answer :
Given $\csc {75^0}$ .
First we find $\sin 75$ .
We know the formula $\sin (a + b) = \sin (a).\cos (b) + \cos (a).\sin (b)$ and 75 can be written as a sum of 45 and 30.
$\Rightarrow \sin 75 = \sin (45 + 30)$
Applying the formula we have.
$= \sin (45).\cos (30) + \cos (45).\sin (30)$
But we know $\sin (45) = \dfrac{1}{{\sqrt 2 }}$ , $\cos (45) = \dfrac{1}{{\sqrt 2 }}$ , $\sin (30) = \dfrac{1}{2}$ and $\cos (30) = \dfrac{{\sqrt 3 }}{2}$ . Substituting we have,
$= \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}$
$= \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}$
Taking LCM and simplifying we have,
$= \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
Thus we have $\Rightarrow \sin 75 = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$ .
We know that $\csc \theta = \dfrac{1}{{\sin \theta }}$ .
$\csc 75 = \dfrac{1}{{\left( {\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}} \right)}}$
Rearranging we have,
$\Rightarrow \csc 75 = \dfrac{{2\sqrt 2 }}{{\left( {\sqrt 3 + 1} \right)}}$ . This is the required answer.
So, the correct answer is “$\dfrac{{2\sqrt 2 }}{{\left( {\sqrt 3 + 1} \right)}}$”.

Note : Remember the formula sine and cosine addition and subtract formula well. In case of sine if we have cosine we will use the formula $\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b)$ . Similarly we have $\sin (a - b) = \sin (a).\cos (b) - \cos (a).\sin (b)$ and $\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b)$ . Depending on the angle we use the required formulas. Also remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.