Answer:
[tex]sin\theta = \sqrt{1-C^2}[/tex]
[tex]tan\theta = \dfrac{\sqrt{1-C^2}}{C}[/tex]
[tex]cot\theta = \dfrac{C}{\sqrt{1-C^2}}[/tex]
[tex]sec\theta = \dfrac{1}{C}}[/tex]
[tex]cosec\theta = \dfrac{1}{\sqrt{1-C^2}}[/tex]
Step-by-step explanation:
Given that:
[tex]\theta[/tex] is an angle in a right angled triangle.
and [tex]cos\theta = C[/tex]
To find:
To draw the triangle and write other five trigonometric functions in terms of C.
Solution:
We know that cosine of an angle is given by the formula:
[tex]cosx =\dfrac{Base}{Hypotenuse}[/tex]
Here, we are given that [tex]cos\theta = C[/tex] OR
[tex]cos\theta = \dfrac{C}{1}[/tex]
i.e. Base = C and Hypotenuse of triangle = 1
Please refer to the right angled triangle as per given statements.
[tex]\triangle PQR[/tex], with base PR = C units
and hypotenuse, QP = 1 unit
[tex]\angle R[/tex] is the right angle.
Let us use pythagorean theorem to find the value of perpendicular.
According to pythagorean theorem:
[tex]\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}[/tex]
[tex]1^{2} = C^{2} + QR^{2}\\\Rightarrow QR = \sqrt {1-C^2}[/tex]
[tex]sin\theta = \dfrac{Perpendicular}{Hypotenuse}\\\Rightarrow sin\theta = \dfrac{\sqrt{1-C^2}}{1}\\\Rightarrow sin\theta = \sqrt{1-C^2}[/tex]
[tex]tan\theta = \dfrac{Perpendicular}{Base}\\\Rightarrow tan\theta = \dfrac{\sqrt{1-C^2}}{C}[/tex]
[tex]cot\theta = \dfrac{Base}{Perpendicular}\\\Rightarrow cot\theta = \dfrac{C}{\sqrt{1-C^2}}[/tex]
[tex]sec\theta = \dfrac{Hypotenuse}{Base}\\\Rightarrow sec\theta = \dfrac{1}{C}}[/tex]
[tex]cosec\theta = \dfrac{Hypotenuse}{Perpendicular}\\\Rightarrow cosec\theta = \dfrac{1}{\sqrt{1-C^2}}[/tex]
