\( (\text{cosec θ}-\text{cot θ})²=\frac{(1-\text{cos θ})}{(1 +\text{cos θ})}\)
L.H.S =\( (\text{cosec θ - cot θ})²\)
\(= \left(\frac{1}{\text{sin θ}} - \frac{\text{cos θ}}{\text{sin θ}}\right)^²\)
\(= \frac{(1 - \text{cos θ})²}{(\text{sin θ})²}\)
\(= \frac{(1 - \text{cos θ})²}{\text{sin² θ}}\)
\(=\frac{ (1 - \text{cos θ})²}{(1 - \text{cos²θ})} \) (By Identity sin A + cos A = 1 Hence, 1 - cos A= sin A)
\(= \frac{(1 - \text{cos θ})²}{ (1 - \text{cos θ})(1 + \text{cos θ})} \) [Using a² - b² = (a + b) (a - b)]
\(=\frac{ (1 -\text{ cos θ})}{(1 + \text{cos θ})}\)
= RHS
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD. 
The relationship between the sides and angles of a right-angle triangle is described by trigonometry functions, sometimes known as circular functions. These trigonometric functions derive the relationship between the angles and sides of a triangle. In trigonometry, there are three primary functions of sine (sin), cosine (cos), tangent (tan). The other three main functions can be derived from the primary functions as cotangent (cot), secant (sec), and cosecant (cosec).
sin x = a/h
cos x = b/h
tan x = a/b
Tan x can also be represented as sin x/cos x
sec x = 1/cosx = h/b
cosec x = 1/sinx = h/a
cot x = 1/tan x = b/a
