# Given sectheta=sqrt2, sintheta=-sqrt2/2 to find the remaining trigonometric function?

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ciumes Share
Apr 13, 2018

If $\sec \theta = \sqrt{2}$ then draw a right triangle in standard position with $x = 1$, $r = \sqrt{2}$. Now, with $\sin \theta$ negative, it will be in QIII or QIV. SInce $x$ is positive, it is in QIV. Now you know $y = - \frac{1}{\sqrt{2}}$
$\sin \theta = \frac{y}{r} = \frac{- \frac{1}{\sqrt{2}}}{\sqrt{2}} = - \frac{1}{2}$
$\cos \theta = \frac{1}{\sec} \theta = \frac{1}{\sqrt{2}}$
$\tan \theta = \frac{y}{x} = - 1$
$\cot \theta = \frac{x}{y} = - 1$
$\csc \theta = \frac{r}{y} = - 2$

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