QUOTIENT RELATION OF TRIGONOMETRIC RATIOS
About "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
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
Quotient relation of trigonometric ratios - Practice problems
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
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