Question

If $\sec \theta = \dfrac{5}{4}$, then the value of $\cos \theta$ is?

Answer 1

In the given problem, we are required to calculate cosine of an angle whose secant is given to us. Such problems require basic knowledge of trigonometric ratios and formulae. We must know the expression of all the trigonometric ratios with reference to a right angled triangle. So, we first assume the sides of a right angled triangle on the basis of the value of the angle. Then, we find all the sides using the Pythagoras theorem. At last, we substitute the values of required sides in the expression of cosine of angle to get to the final answer.

Complete step by step answer:
So, In the given problem, we have to find the value of$\cos \theta$ when $\sec \theta = \dfrac{5}{4}$. To evaluate the value of the required expression, we must keep in mind the formulae of basic trigonometric ratios.
We know that, $\sec \theta = \dfrac{\text{Hypotenuse}}{\text{Base}}$ and $\cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}}$.
So, $\sec \theta = \dfrac{{Hypotenuse}}{{Base}} = \dfrac{5}{4}$
Let the length of the perpendicular of a right angled triangle be $5x$.
Then, length of base of the same triangle $= 4x$.
Hence, $\cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{{4x}}{{5x}}$
Cancelling the common factor in numerator and denominator, we get,
$\Rightarrow \cos \theta = \dfrac{4}{5}$
So, the value of $\cos \theta$ is $\dfrac{4}{5}$ if the value of $\sec \theta = \dfrac{5}{4}$.

Note: The given problem can be solved directly by knowing the fact that the two trigonometric functions, cosine and secant functions are reciprocal functions. This method also provides exposure to the applications of trigonometric identities in various mathematical questions. Care should be taken while handling the calculations. If we had to find any other trigonometric function, then we would have to calculate the length of perpendicular of the same right angled triangle using the Pythagoras theorem.
${\left( \text{Hypotenuse} \right)^2} = {\left( \text{Base} \right)^2} + {\left( \text{Perpendicular} \right)^2}$
Substituting the values of base and hypotenuse, we get,
$\Rightarrow {\left( {5x} \right)^2} = {\left( {4x} \right)^2} + {\left( \text{Perpendicular} \right)^2}$
Computing the squares, we get,
$\Rightarrow 25{x^2} = 16{x^2} + {\left( \text{Perpendicular} \right)^2}$
Shifting the terms in the equation,
$\Rightarrow {\left( \text{Perpendicular} \right)^2} = 25{x^2} - 16{x^2}$
$\Rightarrow {\left( \text{Perpendicular} \right)^2} = 9{x^2}$
Taking square root on both sides of the equation, we get,
$\Rightarrow \left( \text{Perpendicular} \right) = 3x$
So, we get, $\text{Perpendicular} = 3x$
In this way, any trigonometric function: sine, tangent, cotangent, etc. can be found.