What is the value of #tanx + cotx#?

Question

Lincoln Anderson · Accepted Answer

C is the correct answer.

You need to put on a common denominator. This will be #sinxcosx#. #=(sin^2x + cos^2x)/(sinxcosx)# Applying the identity #sin^2x + cos^2x = 1#: #=1/(sinxcosx)# Now, recall that #1/sintheta = csctheta# and #1/costheta = sectheta#. #=cscxsecx# So, C is the answer that corresponds. Hopefully this helps!

Serenity Johnson · Answer

undefined The value of $ \tan(x) + \cot(x) $ can be simplified using the identity $ \cot(x) = \frac{1}{\tan(x)} $. So,

$$
\tan(x) + \cot(x) = \tan(x) + \frac{1}{\tan(x)}
$$

To find a common expression, express both terms with a common denominator:

$$
= \frac{\tan^2(x) + 1}{\tan(x)}
$$

Using the Pythagorean identity $ \tan^2(x) + 1 = \sec^2(x) $, the expression simplifies to:

$$
= \frac{\sec^2(x)}{\tan(x)}
$$

Since $ \sec(x) = \frac{1}{\cos(x)} $ and $ \tan(x) = \frac{\sin(x)}{\cos(x)} $, we can substitute these into the expression:

$$
= \frac{\frac{1}{\cos^2(x)}}{\frac{\sin(x)}{\cos(x)}} = \frac{1}{\sin(x)\cos(x)}
$$

Finally, using the identity $ \sin(2x) = 2\sin(x)\cos(x) $, we can rewrite the expression as:

$$
= \frac{1}{\frac{1}{2}\sin(2x)} = \frac{2}{\sin(2x)}
$$

Thus, the simplified expression for $ \tan(x) + \cot(x) $ is $ \frac{2}{\sin(2x)} $.