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