Step-by-step explanation:
I know that
[tex] \cot {}^{2} (x) + 1 = \csc {}^{2} (x) [/tex]
So These the steps I need to do:
First Apply Reciprocal Identies to find Cot x.
[tex] \frac{1}{ \tan(x) } = \cot(x) [/tex]
So cot x is
[tex] \frac{24}{7} [/tex]
Now, I can squared that result and add 1 to get csc^2(x).
[tex] \frac{576}{49} + 1 = \frac{576}{49} + \frac{49}{49} = \frac{625}{49} [/tex]
Now, I take the principal square root since csc is positve when theta is in the 1st quadrant.
[tex] \frac{ \sqrt{625} }{ \sqrt{49} } = \frac{25}{7} [/tex]
So
[tex] \csc(x) = \frac{25}{7} [/tex]
[tex]cosec{\theta=\frac{25}{7}is\ the\ answer.}[/tex]
[tex]If\ tan\ \theta=\frac{7}{24},0\ degrees < \theta < 90\ degrees[/tex]
[tex]sec\ \theta=\sqrt{1+\tan^2\theta}=\sqrt{1+\frac{49}{576}}=\sqrt{\frac{625}{576}}=\frac{25}{24}[/tex]
[tex]cos\ \theta=\frac{1}{sec\ \theta}=\frac{24}{25}[/tex]
[tex]sin\ \theta=\sqrt{1-\cos^2\theta}=\sqrt{1-(\frac{24}{25})^2}=\sqrt{\frac{49}{625}}=\frac{7}{25}[/tex]
[tex]cosec\ \theta=\frac{1}{sin\ \theta}=\frac{25}{7}[/tex]
[tex]Hence,\ cosec\ \theta=\frac{25}{7}is\ the\ answer.[/tex]
I hope this helps you
:)
