write sec theta in terms of sin theta when theta lies in quadrant III

secθ = 1/cosθ = 1/√(1-sin2θ)

In QIII sec < 0 so we'd have -1/√(1-sin2θ)

Why was the math book sad?

Because it had too many problems!

Now, let's find the expression for sec(theta) in terms of sin(theta) when theta lies in quadrant III.

In quadrant III, both sin(theta) and sec(theta) are negative.

We know that sec(theta) is the reciprocal of cos(theta), so we need to find cos(theta) in terms of sin(theta) first.

Using the Pythagorean identity sin²(theta) + cos²(theta) = 1 and knowing that sin(theta) is negative in quadrant III, we can write it as:

sin²(theta) + (-cos(theta))² = 1

Simplifying this expression, we get:

sin²(theta) + cos²(theta) = 1

Rearranging the terms, we find:

cos²(theta) = 1 - sin²(theta)

Taking the square root of both sides (and remembering that cos(theta) is negative in quadrant III), we have:

cos(theta) = -√(1 - sin²(theta))

Now, we can find sec(theta) by taking the reciprocal of cos(theta):

sec(theta) = -1/√(1 - sin²(theta))

So, in terms of sin(theta), we get:

sec(theta) = -1/√(1 - sin²(theta))

Remember to embrace math humor and not take everything so seriously!

To find $\sec(\theta)$ in terms of $\sin(\theta)$ when $\theta$ lies in Quadrant III, we can utilize the following trigonometric identity:

$\sec(\theta) = \frac{1}{\cos(\theta)}$

In Quadrant III, $\cos(\theta)$ is negative, so we need to account for that by introducing a negative sign.

Now, we know that $\sin^2(\theta) + \cos^2(\theta) = 1$. Rearranging this equation, we get $\cos^2(\theta) = 1 - \sin^2(\theta)$.

Using this expression for $\cos(\theta)$, we can write:
$\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\sqrt{1-\sin^2(\theta)}}$

Therefore, the expression for $\sec(\theta)$ in terms of $\sin(\theta)$ when $\theta$ lies in Quadrant III is $\frac{1}{-\sqrt{1-\sin^2(\theta)}}$.

To write sec(theta) in terms of sin(theta) when theta lies in quadrant III, we can use the formula:

sec(theta) = 1/cos(theta)

Since theta lies in quadrant III, the sine of theta will be negative.

In quadrant III, sin(theta) is negative, so we have: sin(theta) = -abs(sin(theta))

To find cos(theta) in terms of sin(theta), we can use the Pythagorean identity:

sin^2(theta) + cos^2(theta) = 1

Rearranging the equation, we get:

cos^2(theta) = 1 - sin^2(theta)

Taking the square root of both sides and noting that cos(theta) is positive in quadrant III, we have:

cos(theta) = -sqrt(1 - sin^2(theta))

Now, substituting the values of sin(theta) and cos(theta) into the formula for sec(theta), we get:

sec(theta) = 1 / cos(theta)
= 1 / (-sqrt(1 - sin^2(theta)))
= -1 / sqrt(1 - sin^2(theta))

Therefore, sec(theta) in terms of sin(theta) when theta lies in quadrant III is given by:

sec(theta) = -1 / sqrt(1 - sin^2(theta))