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
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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