 # Cosec Cot Formula

A study which deals with the relationship between angles, heights, and lengths of triangles is called Trigonometry. Trigonometry has many applications in the field of engineering, architectural design, astronomy, and physics.

The identities of trigonometry are very useful. There are many fields where these can be applied.

Trigonometry has six main functions, those are a sin, cos, tan, cot, sec, and cosec. All these functions have different formulas. It uses the three sides and angles of a right-angled triangle.

The relationship between the measures of an angle and the length is called trigonometric ratios.

We have hypotenuse, base, and perpendicular in a right angle triangle. And using these three sides, we can find the value of all the six functions.

 Cosec = Hypotenuse / Perpendicular Cot = Base / Perpendicular

## Examples of Cosec Cot formula

Q1) If Sin x = ⅗, find the value of Cosec x?

Cosec x = 1/sinx

= 5/3.

Q2) If Sin x = 5/7, find the value of Cosec x?

Cosec x = 1/sinx

= 7/5

Q.3: Prove that (cosec θ – cot θ)2 = (1 – cos θ)/(1 + cos θ).

Solution:
LHS = (cosec θ – cot θ)2
$=(\frac{1}{sin\ \theta}-\frac{cos\ \theta}{sin\ \theta})^2\\ =(\frac{1-cos\ \theta}{sin\ \theta})^2$ RHS = (1 – cos θ)/(1 + cos θ)
By rationalising the denominator,
$=\frac{1-cos\ \theta}{1+cos\ \theta}\times \frac{1-cos\ \theta}{1-cos\ \theta}\\ =\frac{(1-cos\ \theta)^2}{1-cos^2\theta}\\ =\frac{(1-cos\ \theta)^2}{sin^2\theta}\\ =(\frac{1-cos\ \theta}{sin\ \theta})^2$

Therefore, LHS = RHS
Hence, proved that (cosec θ – cot θ)2 = (1 – cos θ)/(1 + cos θ).

Q.4: Simplify the following expression.
$\sqrt{(1-sin^2A)cosec^2A}$

Solution:
$\sqrt{(1-sin^2A)cosec^2A}\\=\sqrt{cos^2A\times cosec^A}\\=\sqrt{\frac{cos^2A}{sin^2A}}\\ =\sqrt{cot^2A}\\=cot A$