How do you find the remaining trigonometric functions of #theta# given #tantheta=1/2# and #sintheta>0#?
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Given ( \tan(\theta) = \frac{1}{2} ) and ( \sin(\theta) > 0 ), you can find the remaining trigonometric functions using the following steps:

Since ( \tan(\theta) = \frac{1}{2} ), we can use the definition of tangent to find that the opposite side is 1 and the adjacent side is 2.

Using the Pythagorean theorem, we find the hypotenuse to be ( \sqrt{1^2 + 2^2} = \sqrt{5} ).

Since ( \sin(\theta) > 0 ), the sine function is positive in the first and second quadrants. In the first quadrant, ( \sin(\theta) = \frac{1}{\sqrt{5}} ).

Now, you can find the cosine function using the identity ( \cos(\theta) = \frac{1}{\sec(\theta)} ). Since ( \sec(\theta) = \frac{1}{\cos(\theta)} ), ( \cos(\theta) = \frac{2}{\sqrt{5}} ).

Finally, you can find the cosecant, secant, and cotangent functions using the reciprocal identities:
 ( \csc(\theta) = \frac{1}{\sin(\theta)} = \sqrt{5} )
 ( \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{\sqrt{5}}{2} )
 ( \cot(\theta) = \frac{1}{\tan(\theta)} = 2 ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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