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jasenka [17]
3 years ago
7

The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func

tions are reciprocal functions to each other

Mathematics
1 answer:
katrin2010 [14]3 years ago
3 0

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle (\frac{1}{3},\frac{1}4).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A(\frac{1}{3},\frac{1}4)

Let O be the origin i.e. (0,0)

Point B will be (\frac{1}{3},0)

Now, let us consider the right angled triangle \triangle OBA:

Sides:

Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}

Using Pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}

sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}

\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}

cos\angle AOB = \dfrac{Base}{Hypotenuse}

\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}

tan\angle AOB = \dfrac{Perpendicular}{Base}

\Rightarrow tan\angle AOB = \dfrac{3}{4}

cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}

\Rightarrow cosec\angle AOB = \dfrac{5}{3}

sec\angle AOB = \dfrac{Hypotenuse}{Base}

\Rightarrow sec\angle AOB = \dfrac{5}{4}

cot\angle AOB = \dfrac{Base}{Perpendicular}

\Rightarrow cot\angle AOB = \dfrac{4}{3}

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