Find the value for tan 90o, sin 9.
tosec 90° Sec 90
Answers
Answer:
IDKKKKKKKKKKKKKKKKKKKKK
Answer:
I am telling your answer please wait
Explanation:
please mark a brainlist please and follow me
⭐ Answer⭐
Let a rotating line OA rotates about O in the anti-clockwise direction, from initial position to ending position makes an angle ∠XOA = θ. Now a point C is taken on OA and draw CD perpendicular to OX or OX'.
Again another rotating line OB rotates about O in the anti-clockwise direction, from initial position to ending position (OX) makes an angle ∠XOY = 90°; this rotating line now rotates in the clockwise direction, starting from the position (OY) makes an angle ∠YOB = θ.
Now, we can observe that ∠XOB = 90° - θ.
Again a point E is taken on OB such that OC = OE and draw EF perpendicular to
OX or OX'.
Since, ∠YOB = ∠XOA
Therefore, ∠OEF = ∠COD.
Now, from the right-angled ∆EOF and right-angled ∆COD we get, ∠OEF = ∠COD and OE = OC.
Hence, ∆EOF ≅ ∆COD (congruent).
Therefore, FE = OD, OF = DC and OE = OC.
Trigonometrical Ratios of (90° - θ)
4Save
In this diagram FE and OD both are positive. Similarly, OF and DC are both positive.
Trigonometrical Ratios of (90° - θ)
4Save
In this diagram FE and OD both are negative. Similarly, OF and DC are both negative.
Trigonometrical Ratios of (90° - θ)
4Save
In this diagram FE and OD both are negative. Similarly, OF and DC are both negative.
Trigonometrical Ratios of (90° - θ)
4Save
In this diagram FE and OD both are positive. Similarly, OF and DC are both negative.
According to the definition of trigonometric ratio we get,
sin (90° - θ) = FEOE
sin (90° - θ) = ODOC, [FE = OD and OE = OC, since ∆EOF ≅ ∆COD]
sin (90° - θ) = cos θ
cos (90° - θ) = OFOE
cos (90° - θ) = DCOC, [OF = DC and OE = OC, since ∆EOF ≅ ∆COD]
cos (90° - θ) = sin θ
tan (90° - θ) = FEOF
tan (90° - θ) = ODDC, [FE = OD and OF = DC, since ∆EOF ≅ ∆COD]
tan (90° - θ) = cot θ
Similarly, csc (90° - θ) = 1sin(90°−Θ)
csc (90° - θ) = 1cosΘ
csc (90° - θ) = sec θ
sec ( 90° - θ) = 1cos(90°−Θ)
sec (90° - θ) = 1sinΘ
sec (90° - θ) = csc θ
and cot (90° - θ) = 1tan(90°−Θ)
cot (90° - θ) = 1cotΘ
cot (90° - θ) = tan θ