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

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

adjacent side  =  12

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

adjacent side  =  24

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)

cot θ  =  24/7

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