Question

sec theta = 17 ,then tan theta = ?​

Answer 1

$\sin^2 \theta+\cos^2 \theta = 1 \implies \dfrac{\sin^2 \theta}{\cos^2 \theta} + 1 = \dfrac{1}{\cos^2 \theta}$

So $\tan^2 \theta + 1 = \sec^2 \theta$.

 

Given that $\sec^2 \theta = 289$ then $\tan^2 \theta + 1 = 289$.

So $\tan \theta = \pm12\sqrt{2}$.

 

The solutions depend on the angles. Since $\cos \theta > 0$, the values of $\theta$ are in the 1st or the 4th quadrant.

If 1st quadrant: The value of $\tan \theta$ is positive.

If 4th quadrant: The value of $\tan \theta$ is negative.

 

More information:

In the unit circle, which is a circle centered in the origin and radius is 1, the three trigonometric functions are:-

• $\sin \theta = \dfrac{y}{1}$, $\cos \theta = \dfrac{x}{1}$, $\tan \theta = \dfrac{y}{x}$

 

For $\sin \theta$ to be positive $\theta$ should exist in the 1st or the 2nd quadrant.

For $\cos \theta$ to be positive $\theta$ should exist in the 1st or the 4th quadrant.

For $\tan \theta$ to be positive $\theta$ should exist in the 1st or the 3rd quadrant.

[All(1) → Sine(2) → Tangent(3) → Cosine(4)]