So, this is my second post on trigonometry. In this post we're gonna cover the reciprocal and the inverse Trigonometric functions. If you haven't seen my first post you should definitely view it as it covers the basics of Trigonometry

## The Reciprocal Trigonometric Functions

The reciprocal Trigonometric function of Sine is Cosecant, of Cosine is Secant & for Tangent it is Cotangent.

- Cosecant (Csc θ = 1/Sin θ) or (Hypotenuse/Opposite)
- Secant (Sec θ = 1/Cos θ) or (Hypotenuse/Adjacent)
- Cotangent (Cot θ = 1/Tan θ) or (Adjacent/Opposite)

We can also represent Tan θ in another way. As Tan θ = opposite/adjacent

& Sin θ = opposite/hypotenuse

& Cos θ = adjacent/hypotenuse

∴ Tan θ = Sin θ/Cos θ (The hypotenuses cancel out)

As Cot θ = 1/Tan θ

So, we can also represent Cot θ as Cos θ/Sin θ.

## The Inverse Trigonometric Functions

In trigonometry the inverse trigonometric functions sin

^{-1 }, cos^{-1}, tan^{-1}, csc^{-1}, sec^{-1}, cot^{-1}(aka cyclometric functions) are the inverse functions of sin, cos, tan, csc, sec, cot respectively.
This means that the sin

^{-1}of a value, say x would be the angle which gives x when its sine is taken
If you cant understand this, I'll give some examples which would make you understand the concept

sin

^{-1}(1/2) = ? (This means that which angle when taken the sin of, gives the value as 1/2)
So, sin

^{-1}(1/2) =30° (sin 30 = 1/2)## Example

Heres an example that can make you see the use of the inverse trigonometric functions

There's an alien standing on the top of Burj khalifa (828 meters), and there's a MIB agent standing on the roof of another building 8 meters long and the distance between the base of the burj khalifa and the building where the agent is standing is 100 meters. Find the angle at which the agent should shoot his Ray gun to vaporize the alien. Round off your answer to the nearest hundredths.

Solution - Lets first construct the triangle which joins the agent, the alien and the point 8m above the base of Burj khalifa. Lets label the required angle as θ

The opposite side to theta will be 820 m (total height of building - the height at which the agent is standing)

We know 2 sides - The adjacent and the opposite

Which trigonometric function deals with opposite and adjacent? Well, we know that Tan θ = opp/adj

_{}

^{}

Tan θ = 82/10 = 41/5 = 8.2

Tan

^{-1}(Tan θ) = θ
⇒Tan

use calculator to figure Tan

θ = 83.0470425°

θ ≈ 83.05°

So, the agent has to shoot his gun roughly at an angle of 83.05° to hit the alien

I hope you liked the post, if you have any doubt, leave it in the comments section below.

In the next post we will move on to trigonometry with general triangles and also some special right triangles.

Stay tuned for more :D

^{-1}(8.2) = θuse calculator to figure Tan

^{-1}(8.2) ,θ = 83.0470425°

θ ≈ 83.05°

So, the agent has to shoot his gun roughly at an angle of 83.05° to hit the alien

I hope you liked the post, if you have any doubt, leave it in the comments section below.

In the next post we will move on to trigonometry with general triangles and also some special right triangles.

Stay tuned for more :D