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Find the exact value of each of the remaining trigonometric functions of theta.

\[cos\theta=\frac{24}{25}\ ,\ {270} ^\circ<\theta<{360} ^\circ\]

– Part (a) – $sin\theta=?$

– Part (b) – $tan\theta=?$

– Part (c) – $sec\theta=?$

– Part (d) – $csc\theta=?$

– Part (e) – $cot\theta=?$

The aim of the article is to find the value of trigonometric functions of the Right-Angled triangleThe basic concept behind this article is the Right-Angled Triangle and the Pythagorean Identity.

A triangle is called Right-Angled Triangle if it contains one internal angle of ${90}^\circ$ and the other two internal angles sum up with the Right Angle to complete ${180}^\circ$. The horizontal side of the Right Angle is called the Adjacent, and the Vertical Side is called the Opposite.

The Pythagorean Identity for the Right-Angled Triangle is expressed as follows:

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

This is true for all values of angles $\theta$.

Expert Answer

Given that:

\[cos\theta=\frac{24}{25}\ ,\ {270}^\circ<\theta<{360}^\circ\]

The given range of angle represents that the angle $\theta$ lies in the $4^{th}$ quadrant.

Part (a) – $sin\theta=?$

As per the Pythagorean Identity, we know that:

\[\sin^2\theta+{\ \cos}^2\theta=1\]

\[sin\theta\ =\ \sqrt{1-\cos^2\theta}\]

Substituting the value of $cos\theta=\dfrac{24}{25}$:

\[sin\theta=\sqrt{1-\left(\frac{24}{25}\right)^2}\]

\[sin\theta=\sqrt{\frac{625-576}{625}}\]

\[sin\theta=\sqrt{\frac{49}{625}}\]

\[sin\theta=\pm\frac{7}{25}\]

Since the angle $\theta$ lies in $4^{th}$ quadrant, the $sine$ function will be negative:

\[sin\theta=-\frac{7}{25}\]

Part (b) – $tan\theta=?$

We know that for the Right-Angled Triangle:

\[tan\theta=\frac{sin\theta}{cos\theta}\]

Substituting the value of $sin\theta$ and $cos\theta$ in the above equation:

\[tan\theta=\frac{-\dfrac{7}{25}}{\dfrac{24}{25}}\]

\[tan\theta=-\frac{7}{25}\times\frac{25}{24}\]

\[tan\theta=-\frac{7}{24}\]

Part (c) – $sec\theta=?$

We know that for the Right-Angled Triangle:

\[sec\theta=\frac{1}{cos\theta}\]

Substituting the value $cos\theta$ in the above equation:

\[sec\theta=\frac{1}{\dfrac{24}{25}}\]

\[sec\theta=\frac{25}{24}\]

Part (d) – $csc\theta=?$

We know that for the Right-Angled Triangle:

\[csc\theta=\frac{1}{sin\theta}\]

Substituting the value $sin\theta$ in the above equation:

\[csc\theta=\frac{1}{-\dfrac{7}{25}}\]

\[csc\theta=-\frac{25}{7}\]

Part (e) – $cot\theta=?$

We know that for the Right-Angled Triangle:

\[cot\theta=\frac{1}{tan\theta}\]

Substituting the value $tan\ \theta$ in the above equation:

\[cot\theta=\frac{1}{-\dfrac{7}{24}}\]

\[cot\theta=-\frac{24}{7}\]

Numerical Result

Part (a) – $sin\ \theta\ =\ -\ \dfrac{7}{25}$

Part (b) – $tan\ \theta\ =\ -\ \dfrac{7}{24}$

Part (c) – $sec\ \theta\ =\ \dfrac{25}{24}$

Part (d) – $csc\ \theta\ =\ -\ \dfrac{25}{7}$

Part (e) – $cot\ \theta\ =\ -\ \dfrac{24}{7}$

Example

Calculate the value for the following trigonometric functions if:

\[cos\ \theta\ =\ \frac{3}{5}\ ,\ {90}^\circ\ <\ \theta\ <\ {180}^\circ \]

Part (a) – $sin\ \theta\ =\ ?$

Part (b) – $tan\ \theta\ =\ ?$

Solution

Given that:

\[cos\ \theta\ =\ \frac{3}{5}\ ,\ {90}^\circ\ <\ \theta\ <{\ 180}^\circ \]

The given range of angle represents that the angle $\theta$ lies in the $2^{nd}$ quadrant.

Part (a) – $sin\ \theta\ =\ ?$

As per the Pythagorean Identity, we know that:

\[\sin^2\ \theta+{\ \cos}^2\ \theta\ =\ 1 \]

\[sin\theta\ =\ \sqrt{1\ -{\cos}^2\ \theta} \]

Substituting the value of $cos\ \theta\ =\ \dfrac{3}{5}$:

\[sin\ \theta\ =\ \sqrt{1\ -{\ \left(\frac{3}{5}\right)}^2} \]

\[sin\ \theta\ =\ \sqrt{\frac{25\ -\ 9}{25}} \]

\[sin\ \theta\ =\ \sqrt{\frac{16}{25}} \]

\[sin\ \theta\ =\ \pm\ \frac{4}{5} \]

Since the angle $\theta$ lies in the $2^{nd}$ quadrant, the $sine$ function will be positive:

\[sin\ \theta\ =\ \ \frac{4}{5} \]

Part (b) – $tan\ \theta\ =\ ?$

We know that for the Right-Angled Triangle:

\[tan\ \theta\ =\ \frac{sin\ \theta}{cos\ \theta} \]

Substituting the value of $sin\ \theta$ and $cos\ \theta$ in the above equation:

\[tan\ \theta\ =\ \frac{\ \dfrac{4}{5}}{\dfrac{3}{5}} \]

\[tan\ \theta\ =\ \frac{4}{5}\ \times\ \frac{5}{3} \]

\[tan\ \theta\ =\ \frac{4}{3} \]

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