Find the exact value of secθ if cscθ=-4/3 and the terminal side of θ lies in Quadrant III.

1 answer:

6 0

Quadrant is III so, sign would be negative for Sec.
Now, we know, csc = H / P
So, H = 4, P = 3
Calculate for B,
B² = H² - P²
B² = 4² - 3²
B² = 16 - 9
B = √7

We know, sec Ф = H / B = 4 / √7
We can write it as: -4√7 / 7 [ -ve sign for 3rd quadrant ]

In short, Your Answer would be Option B

Hope this helps!

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What are the coordinates of the point on the directed line segment from (-6, -6)(−6,−6) to (9, -1)(9,−1) that partitions the seg

GrogVix [38]

Answer:

(0, -4)

Step-by-step explanation:

The coordinates of the points from which the directed line segment extends = (-6, -6) to (9, -1)

The ratio the required point partitions the line = 2 to 3

The formula for finding the coordinate of a point that partitions a line AB into a ratio 'a' to 'b', where the coordinates of, A = (x₁, y₁) and B = (x₂, y₂) is given as follows;

$\left(\dfrac{a}{a + b} \times (x_1 - x_2)+ x_1, \ \dfrac{a}{a + b} \times (y_1 - y_2)+ y_1 \right)$

Therefore, the required point is located as follows;

$\left(\dfrac{2}{2 + 3} \times (9 - (-6))+ (-6), \ \dfrac{2}{2 + 3}\times (-1 - (-6))+ (-6) \right) = (0, -4)$

The coordinates of the point is (0, -4)

6 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
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and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$