Question

prove that .. cos(90-theta)*sin(90-theta)/tan(90-theta)=sin^theta

Answer 1

cos (90-theta)*sin(90-theta)
sin theta*cos theta (complementry angles)
sin theta * cos theta =1

Answer 2

▶▶▶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) = FEOEFEOE

sin (90° + θ) = ODOCODOC, [FE = OD and OE = OC, since ∆ OCD ≅ ∆ OEF]

sin (90° + θ) = cos θ


▶▶▶cos (90° + θ) = OFOEOFOE

cos (90° + θ) = −DCOC−DCOC, [OF = -DC and OE = OC, since ∆ OCD ≅ ∆ OEF]

cos (90° + θ) = - sin θ.


▶▶▶tan (90° + θ) = FEOFFEOF

tan (90° + θ) = OD−DCOD−DC, [FE = OD and OF = - DC, since ∆ OCD ≅ ∆ OEF]

tan (90° + θ) = - cot θ.


▶▶▶▶▶Similarly, csc (90° + θ) = 1sin(90°+Θ)1sin(90°+Θ)

csc (90° + θ) =  1cosΘ1cosΘ

csc (90° + θ) = sec θ.


sec (90° + θ) = 1cos(90°+Θ)1cos(90°+Θ) 

sec (90° + θ) =  1−sinΘ1−sinΘ

sec (90° + θ) = - csc θ.


▶▶▶and cot (90° + θ) = 1tan(90°+Θ)1tan(90°+Θ)

cot (90° + θ) = 1−cotΘ1−cotΘ

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

            = 1√21√2


2. Find the value of tan 150°.

▶▶▶▶▶Solution:

tan 150° = tan (90 + 60)°

            = - cot 60°; since we know, tan (90° + θ) = - cot θ

            = 1√31√3