# QUOTIENT RELATION OF TRIGONOMETRIC RATIOS

Quotient Relation of Trigonometric Ratios :

When we divide the trigonometric ratio sinθ by cosθ, the quotient is tanθ.

When we divide the trigonometric ratio cosθ by sinθ, the quotient is cotθ.

When we divide the trigonometric ratio cscθ by secθ, the quotient is cotθ.

When we divide the trigonometric ratio secθ by cscθ, the quotient is tanθ.

The above said divisions are called as quotient relation of trigonometric ratios.

## Quotient Relation of Trigonometric Ratios In the triangle above, according SOHCAHTOA, we have

sin θ  =  opposite side / hypotenuse  =  BC / AC

cos θ  =  adjacent side / hypotenuse  =  AB / AC

Now, let us divide sin θ by cos θ.

sin θ / cos θ  =  (BC/AC) ÷ (AB/AC)

sin θ / cos θ  =  (BC/AC)  (AC/AB)

sin θ / cos θ  =  BC /AB

sin θ / cos θ  =  tan θ

(Because, tanθ = opposite side / adjacent side  =  BC / AB)

Therefore,

sin θ / cos θ  =  tan θ

Now, let us divide cos θ by sin θ.

cos θ / sin θ  =  (AB/AC) ÷ (BC/AC)

cos θ / sin θ  =  (AB/AC)  (AC/BC)

cos θ / sin θ  =  AB / BC

cos θ / sin θ  =  cot θ

(Because, cotθ = adjacent side / opposite side  =  AB / BC)

Therefore,

cos θ / sin θ  =  cot θ

csc θ  =  1 / sin θ  =  AC / BC

sec θ  =  1 / cos θ  =  AC / AB

Now, let us divide csc θ by sec θ.

csc θ / sec θ  =  (AC/BC) ÷ (AC/AB)

csc θ / sec θ  =  (AC/BC)  (AB/AC)

csc θ / sec θ  =  AB / BC

csc θ / sec θ  =  cot θ

(Because, cotθ = adjacent side / opposite side  =  AB / BC)

Therefore,

csc θ / sec θ  =  cot θ

Now, let us divide sec θ by csc θ.

sec θ / csc θ  =  (AC/AB) ÷ (AC/BC)

sec θ / csc θ  =  (AC/AB)  (BC/AC)

sec θ / csc θ  =  BC / AB

sec θ / csc θ  =  tan θ

(Because, tanθ = opposite side/adjacent side  =  BC / AB)

Therefore,

sec θ / csc θ  =  tan θ

## Quotient Relation of Trigonometric Ratios - Practice Problems

Problem 1 :

In the right triangle PQR shown below, find the value of sin θ and cos θ. Using them, find the value of tan θ and cot θ. Solution :

From the right triangle shown above,

opposite side  =  5

hypotenuse  =  13

Therefore,

sin θ  =  PQ/RQ  =  5/13

cos θ  =  PR/RQ  =  12/13

tan θ  =  sin θ / cos θ  =  (5/13) ÷ (12/13)

tan θ  =  (5/13)  (13/12)

tan θ  =  5/12

cot θ  =  cos θ / sin θ  =  (12/13) ÷ (5/13)

cot θ  =  (12/13)  (13/5)

cot θ  =  12/5

Problem 2 :

From the figure given below, find the value of sin θ and cos θ. Using them, find the value of tan θ and cot θ. Solution :

From the figure given above, AC = 24 and BC = 7.

By Pythagorean theorem,

AB2  =  BC2 + CA2

AB2  =  72 + 242

AB2  =  49 + 576

AB²  =  49 + 576

AB2  =  625

AB2  =  252

AB  =  25

Now, we can use the three sides to find the six trigonometric ratios of angle θ.

Therefore,

opposite side  =  7

hypotenuse  =  25

Therefore,

sin θ  =  BC/AB  =  7/25

cos θ  =  AC/AB  =  24/25

tan θ  =  sin θ / cos θ  =  (7/25) ÷ (24/25)

tan θ  =  (7/25)  (25/24)

tanθ  =  7/24

sin θ  =  BC/AB  =  7/25

cos θ  =  AC/AB  =  24/25

cot θ  =  cos θ / sin θ  =  (24/25) ÷ (7/25)

cot θ  =  (24/25)  (25/7) After having gone through the stuff given above, we hope that the students would have understood the quotient relation of trigonometric ratios.

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