**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.

In the triangle above, according SOHCAHTOA, we have

sinθ = opposite side / hypotenuse = BC / AC

cosθ = adjacent side / hypotenuse = AB / AC

Now, let us divide sinθ by cosθ

sinθ / cosθ = (BC/AC) ÷ (AB/AC)

sinθ / cosθ = (BC/AC) x (AC/AB)

sinθ / cosθ = BC /AB = tanθ

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

Therefore,

sinθ / cosθ = tanθ

Now, let us divide cosθ by sinθ

cosθ / sinθ = (AB/AC) ÷ (BC/AC)

cosθ / sinθ = (AB/AC) x (AC/BC)

cosθ / sinθ = AB / BC = cotθ

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

Therefore,

cosθ / sinθ = cotθ

cscθ = 1 / sinθ = AC / BC

secθ = 1 / cosθ = AC / AB

Now, let us divide cscθ by secθ

cscθ / secθ = (AC/BC) ÷ (AC/AB)

cscθ / secθ = (AC/BC) x (AB/AC)

cscθ / secθ = AB / BC = cotθ

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

Therefore,

cscθ / secθ = cotθ

Now, let us divide secθ by cscθ

secθ / cscθ = (AC/AB) ÷ (AC/BC)

secθ / cscθ = (AC/AB) x (BC/AC)

secθ / cscθ = BC / AB = tanθ

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

Therefore,

secθ / cscθ = tanθ

**Problem 1 :**

In the right triangle PQR given below, find the value of sinθ and cosθ. Using them, find the value of tanθ and cotθ

**Solution :**

From the figure given above,

opposite side = 5

adjacent side = 12

hypotenuse = 13

Therefore,

sinθ = PQ/RQ = 5/13

cosθ = PR/RQ = 12/13

tanθ = sinθ / cosθ = (5/13) ÷ (12/13)

tanθ = (5/13) x (13/12)

**tanθ = 5/12**

cotθ = cosθ / sinθ = (12/13) ÷ (5/13)

cotθ = (12/13) x (13/5)

**cotθ = 12/5**

**Problem 2 :**

From the figure given below, find the value of sinθ and cosθ. Using them, find the value of tanθ and cotθ

**Solution : **

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

By Pythagorean theorem,

AB² = BC² + CA²

AB² = 7² + 24²

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,

sinθ = BC/AB = 7/25

cosθ = AC/AB = 24/25

tanθ = sinθ / cosθ = (7/25) ÷ (24/25)

tanθ = (7/25) x (25/24)

**tanθ = 7/24**

sinθ = BC/AB = 7/25

cosθ = AC/AB = 24/25

cotθ = cosθ / sinθ = (24/25) ÷ (7/25)

cotθ = (24/25) x (25/7)

**cotθ = 24/7**

After having gone through the stuff given above, we hope that the students would have understood "Quotient relation of trigonometric ratios"

If you want to know more about "Quotient relation of trigonometric ratios", please click here

If you need any other stuff in math, please use our google custom search here.

HTML Comment Box is loading comments...

**WORD PROBLEMS**

**HCF and LCM word problems**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**

**Sum of all three four digit numbers formed using 0, 1, 2, 3**

**Sum of all three four digit numbers formed using 1, 2, 5, 6**