Find the value of


Find the value of

$\sec \left(-1470^{\circ}\right)$



To find: Value of sec (-1470°)

We have,

$\sec \left(-1470^{\circ}\right)=\sec \left(1470^{\circ}\right)$

$[\because \sec (-\theta)=\sec \theta]$

$=\sec \left[90^{\circ} \times 16+30^{\circ}\right]$

Clearly, $1470^{\circ}$ is in Ist Quadrant and the multiple of $90^{\circ}$ is even

$=\sec 30^{\circ}$

$=\frac{2}{\sqrt{3}}\left[\because \sec 30^{\circ}=\frac{2}{\sqrt{3}}\right]$


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