prove that .. cos(90-theta)*sin(90-theta)/tan(90-theta)=sin^theta
Attachments:
Answers
Answered by
0
cos (90-theta)*sin(90-theta)
sin theta*cos theta (complementry angles)
sin theta * cos theta =1
sin theta*cos theta (complementry angles)
sin theta * cos theta =1
alfaazahmed25:
sorry this is not the answer
Answered by
0
▶▶▶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
▶▶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
Similar questions