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 anticlockwise 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 rightangled ∆ 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
= \(\frac{1}{√3}\)
11 and 12 Grade Math
From Trigonometrical Ratios of (90° + θ) to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.