Trigonometric Ratios

Fundamental relations between the trigonometric ratios of an angle:

To know the **relations between the** **trigonometric ratios**** **from the above figure, we see;

sin θ = perpendicular/hypotenuse = MP/PO and

cosec θ = hypotenuse/perpendicular = PO/MP

**It is clear that one
is the reciprocal of the other.**

So, sin θ = 1/cosec θ and

cosec θ = 1/sin θ **……….
(a)**

Again, cos θ = base/hypotenuse = OM/OP and

sec θ = hypotenuse/ base = OP/OM

**One is reciprocal of
the other.**

That is, cos θ = 1/sec θ and sec θ = 1/cos θ **………. (b)**

So, tan θ = perpendicular/base = MP/OM and cot θ = base/perpendicular = OM/MP

tan θ = 1/cot θ and cot θ = 1/tan θ **………. (c)**

Moreover, sin θ/cos θ = (MP/OP) ÷ (OM/OP) = (MP/OP) × (OP/OM) = MP/OM = tan θ

Therefore, sin θ/cos θ = tan θ **………. (d)**

and cos θ/sin θ = (OM/OP) ÷ (MP/OP) = (OM/OP) × (OP/MP) = OM/MP = cot θ

Therefore, cos θ/sin θ = cot θ **………. (e)**

Cos θ = OM/OP

Tan θ = PM/OM

Csc θ = OP/PM

Sec θ = OP/OM

Cot θ = OM/PM

Now from the right-angled triangle POM we get;

PM

Dividing both sides by OP

PM

or, (PM/OP)

or,

Again, dividing both sides of (i) by OM

PM

or, (PM/OM)

or,

Finally, dividing both of (i) by PM

PM

or, 1 + (OM/PM)

or,

(i)

(ii)

(i)

(ii)

(i)

(ii)

This is how the ratios are related to show that **one is the reciprocal of the other** according to the relations between the trigonometric ratios.

**Relations Between the Trigonometric Ratios**

**Problems on Trigonometric Ratios**

**Reciprocal Relations of Trigonometric Ratios**

**Problems on Trigonometric Identities**

**Elimination of Trigonometric Ratios**** **

**Eliminate Theta between the equations**

**Verify Trigonometric Identities**** **

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