What is the relation among all the trigonometrical ratios of (– θ)?
In trigonometrical ratios of angles ( θ) we will find the relation between all six trigonometrical ratios.
Let a rotating line OA rotates about O in the anticlockwise direction. From initial position to ending position OA make an angle ∠XOA = θ.
Again a rotating line OA rotates about O in the clockwise direction and makes an angle ∠XOB having magnitude equal to ∠XOA.
Then we get, ∠XOB =  θ. Observe the diagram 1 and 4 to take a point C on OA and draw CD perpendicular to OX. Or we can also observe the diagram 2 and 3 where CD perpendicular to OX'. Let produce CD to intersect OB at E. Now, from the ∆ COD and ∆ EOD we get ∠COD = ∠EOD (same magnitude), ∠ODC = ∠ODE and OD is common.
Therefore, ∆ COD
≅ ∆ EOD (congruent)
Therefore, according to the rules of trigonometric sign we get,
ED =  CD and OE = OC.
Again according to the definition of trigonometric ratios,
sin ( θ) = \(\frac{ED}{OE}\)
sin ( θ) = \(\frac{ CD}{OC}\), [ED = CD and OE = OC since, ∆ COD ≅ ∆ EOD]
sin ( θ) =  sin θ
again, cos ( θ) = \(\frac{OD}{OE}\)
cos ( θ) = \(\frac{OD}{OC}\), [OE = OC since, ∆ COD ≅ ∆ EOD]
cos ( θ) = cos θ
again, tan ( θ) = \(\frac{ED}{OD}\)
tan ( θ) = \(\frac{ CD}{OD}\), [ED = CD since, ∆ COD ≅ ∆ EOD]
tan ( θ) =  tan θ.
similarly, csc ( θ) = \(\frac{1}{sin ( \Theta)}\)
csc ( θ) = \(\frac{1}{ sin \Theta}\)
csc ( θ) =  csc θ.
again, sec ( θ) = \(\frac{1}{cos ( \Theta)}\)
sec ( θ) = \(\frac{1}{cos \Theta}\)
sec ( θ) = sec θ.
And again, cot ( θ) = \(\frac{1}{tan ( \Theta)}\)
cot ( θ) = \(\frac{1}{ tan \Theta}\)
cot ( θ) =  cot θ.
Solved example:
1. Find the value of sin ( 45)°.
Solution:
sin ( 45)° =  sin 45°; since we know sin ( θ) =  sin θ
= \(\frac{1}{√2}\)
2. Find the value of sec ( 60)°.
Solution:
sec ( 60)° = sec 60°; since we know sec ( θ) = sec θ
= 2
3. Find the value of cot ( 90)°.
Solution:
cot ( 90)° =  tan 90°; since we know cot ( θ) =  tan θ
= 0
11 and 12 Grade Math
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