In trigonometrical ratios of angles (90° + θ) us will discover the relation in between all 6 trigonometrical ratios.
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Let a rotating line OA rotates about O in the anti-clockwise direction, from initial position to finishing position provides an angle ∠XOA = θ again the exact same rotating heat rotates in the same direction and makes an angle ∠AOB =90°.
Take a point C top top OA and also draw CD perpendicular come OX or OX’.
Again, take it a allude E top top OB such the OE = OC and draw EF perpendicular to OX or OX’. Indigenous the right-angled ∆ OCD and also ∆ OEF us get,
∠COD = ∠OEF
and OC = OE.
Therefore, ∆ OCD ≅ ∆ OEF (congruent).
Therefore according to the meaning of trigonometric sign, the = - DC, FE = OD and also OE = OC
We observe that in chart 1 and 4 OF and also DC room opposite signs and also FE, OD room either both positive. Again us observe that in diagram 2 and also 3 OF and also DC are opposite signs and FE, OD are both negative.
According to the an interpretation of trigonometric ratio we get,
sin (90° + θ) = (fracFEOE)
sin (90° + θ) = (fracODOC),
sin (90° + θ) = cos θ
cos (90° + θ) = (fracOFOE)
cos (90° + θ) = (frac- DCOC),
cos (90° + θ) = - sin θ.
tan (90° + θ) = (fracFEOF)
tan (90° + θ) = (fracOD- DC),
tan (90° + θ) = - cot θ.
Similarly, csc (90° + θ) = (frac1sin (90° + Theta))
csc (90° + θ) = (frac1cos Theta)
csc (90° + θ) = sec θ.
sec (90° + θ) = (frac1cos (90° + Theta))
sec (90° + θ) = (frac1- sin Theta)
sec (90° + θ) = - csc θ.
and cot (90° + θ) = (frac1tan (90° + Theta))
cot (90° + θ) = (frac1- cot Theta)
cot (90° + θ) = - tan θ.
Solved examples:
1. find the worth of sin 135°.
Solution:
sin 135° = sin (90 + 45)°
= cos 45°; because we know, sin (90° + θ) = cos θ
= (frac1√2)
2. discover the worth of tan 150°.
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Solution:
tan 150° = tan (90 + 60)°
= - cot 60°; due to the fact that we know, tan (90° + θ) = - cot θ
= (frac1√3)
● Trigonometric Functions
11 and 12 class Math indigenous Trigonometrical Ratios the (90° + θ) to home PAGE