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Levart [38]
3 years ago
11

If sin theta = -4/5 and theta is in quadrant 4, the value of sec theta

Mathematics
2 answers:
katovenus [111]3 years ago
4 0
Sec theta would be 5/3
Citrus2011 [14]3 years ago
4 0

Answer:

\dfrac{5}{3}

Step-by-step explanation:

We know that \sec{\theta}=\dfrac{1}{\cos{\theta}}. On the other hand,

we can use the equality

\sin^2 \theta + \cos^2 \theta=1

to find the absolute value of \cos \theta. It holds that

\left[\dfrac{4}{5}\right]^2+\cos^2 \theta = 1 \Rightarrow \\\\\\\quad \cos^2 \theta = 1-\dfrac{16}{25}=\dfrac{9}{25} \Rightarrow\\ \\\cos \theta = \pm \dfrac{3}{5}

Now, in the quadrant 4 the values of cosine is positive for all theta, hence

\cos{\theta}=\dfrac{3}{5} \quad \Rightarrow  \quad \sec \theta=\dfrac{1}{\cos\theta}=\dfrac{5}{3}

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It's not difficult to compute the values of A and B directly:

A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}
A=\tan^{-1}(\sin\theta)-\dfrac\pi4

B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt
B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}
B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2

Let's assume 0, so that |\csc\theta|=\csc\theta.

Now,

\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}
\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}
\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)
\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in A and B.
3 0
3 years ago
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Answer:

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Step-by-step explanation:

I hope this helps you in any way.Have a great day :)

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