Question

If $\sec \phi =\dfrac{5}{4}$ and ${{0}^{\circ }}<\phi <{{90}^{\circ }}$. How do you find $\sec 2\phi$ ?

Answer 1

We explain the function $\sec \phi =\dfrac{5}{4}$ and the quadrant value for the angle $\phi$. We express the identity functions of other ratio of cos with ratio of sec. It’s given that $\sec \phi =\dfrac{5}{4}$ and ${{0}^{\circ }}<\phi <{{90}^{\circ }}$ which means the angle is in quadrant I. Thereafter we put the value to find the value of each of the remaining trigonometric function. We also use the multiple angle formula of $\cos 2\phi =2{{\cos }^{2}}\phi -1$.

Complete step by step answer:
It’s given that $\sec \phi =\dfrac{5}{4}$, $\phi$ being in quadrant I as ${{0}^{\circ }}<\phi <{{90}^{\circ }}$. In that quadrant all ratios are positive.
We can find the value of $\cos \phi$ from the relation of $\left( \cos x \right)=\dfrac{1}{\sec x}$.
The value of $\cos \phi$ in quadrant I will be positive.So,
$\left( \cos \phi \right)=\dfrac{1}{\sec \phi }=\dfrac{4}{5}$.
Now we use the multiple angle formula of $\cos 2\phi =2{{\cos }^{2}}\phi -1$ for $\cos 2\phi$.So,
$\cos 2\phi =2{{\left( \dfrac{4}{5} \right)}^{2}}-1\\\ \Rightarrow\cos 2\phi =\dfrac{32}{25}-1\\\ \Rightarrow\cos 2\phi =\dfrac{7}{25}$
We use the relation $\left( \cos x \right)=\dfrac{1}{\sec x}$ again to find the value of $\sec 2\phi$ from $\cos 2\phi$.
Therefore,

\therefore\left( \sec 2\phi \right)=\dfrac{25}{7}$$ **Hence, the exact value of $$\sec 2\phi $$ is $$\dfrac{25}{7}$$.** **Note:** In addition to the reciprocal relationships of certain trigonometric functions, two other types of relationships exist. These relationships, known as trigonometric identities, include functional relationships and Pythagorean relationships.