What is the relation among all the trigonometrical ratios of (90° + θ)?
In trigonometrical ratios of angles (90° + θ) we will find the relation between all six trigonometrical ratios.
Let a rotating line OA rotates about O in the anti-clockwise direction, from initial position to ending position makes an angle ∠XOA = θ again the same rotating line rotates in the same direction and makes an angle ∠AOB =90°.
Therefore we see that, ∠XOB = 90° + θ.
Take a point C on OA and draw CD perpendicular to OX or OX’.
Again, take a point E on OB such that OE = OC and draw EF perpendicular to OX or OX’. From the right-angled ∆ OCD and ∆ OEF we get,
∠COD = ∠OEF [since OB ⊥ OA]
and OC = OE.
Therefore, ∆ OCD ≅ ∆ OEF (congruent).
Therefore according to the definition of trigonometric sign, OF = - DC, FE = OD and OE = OC
We observe that in diagram 1 and 4 OF and DC are opposite signs and FE, OD are either both positive. Again we observe that in diagram 2 and 3 OF and DC are opposite signs and FE, OD are both negative.
According to the definition of trigonometric ratio we get,
sin (90° + θ) = \(\frac{FE}{OE}\)
sin (90° + θ) = \(\frac{OD}{OC}\), [FE = OD and OE = OC, since ∆ OCD ≅ ∆ OEF]
sin (90° + θ) = cos θ
cos (90° + θ) = \(\frac{OF}{OE}\)
cos (90° + θ) = \(\frac{- DC}{OC}\), [OF = -DC and OE = OC, since ∆ OCD ≅ ∆ OEF]
cos (90° + θ) = - sin θ.
tan (90° + θ) = \(\frac{FE}{OF}\)
tan (90° + θ) = \(\frac{OD}{- DC}\), [FE = OD and OF = - DC, since ∆ OCD ≅ ∆ OEF]
tan (90° + θ) = - cot θ.
Similarly, csc (90° + θ) = \(\frac{1}{sin (90° + \Theta)}\)
csc (90° + θ) = \(\frac{1}{cos \Theta}\)
csc (90° + θ) = sec θ.
sec (90° + θ) = \(\frac{1}{cos (90° + \Theta)}\)
sec (90° + θ) = \(\frac{1}{- sin \Theta}\)
sec (90° + θ) = - csc θ.
and cot (90° + θ) = \(\frac{1}{tan (90° + \Theta)}\)
cot (90° + θ) = \(\frac{1}{- cot \Theta}\)
cot (90° + θ) = - tan θ.
Solved examples:
1. Find the value of sin 135°.
Solution:
sin 135° = sin (90 + 45)°
= cos 45°; since we know, sin (90° + θ) = cos θ
= \(\frac{1}{√2}\)
2. Find the value of tan 150°.
Solution:
tan 150° = tan (90 + 60)°
= - cot 60°; since we know, tan (90° + θ) = - cot θ
= \(\frac{1}{√3}\)
● Trigonometric Functions
11 and 12 Grade Math
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