Question

write the following rational number with denominator (-30) 1) 2/3 2) -3/5​

Answer 1

Answer:

Step-by-step explanation:

What is the relation among all the trigonometrical ratios of (90° - θ)?

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. 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)

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In this diagram FE and OD both are positive. Similarly, OF and DC are both positive.

Trigonometrical Ratios of (90° - θ)

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In this diagram FE and OD both are negative. Similarly, OF and DC are both negative.

Trigonometrical Ratios of (90° - θ)

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In this diagram FE and OD both are negative. Similarly, OF and DC are both negative.

Trigonometrical Ratios of (90° - θ)

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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

Solved examples:

1. Find the value of cos 30°.

Solution:

cos 30° = sin (90 - 60)°

           = sin 60°; since we know, cos (90° - θ) = sin θ

             = √32

2. Find the value of csc 90°.

Solution:

csc 90° = csc (90 - 0)°

           = sec 0°; since we know, csc (90° - θ) = sec θ

             = 1