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Trigonometric Ratios

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Trigonometric Ratios:

As we know, in the earlier chapter that right angled triangle consists of three sides i.e ; Perpendicular , Base and Hypotenuse then as we talk about ratios 6 ratios each of them obtained from these 3 sides they are :

  1. \(\frac{p} {h}\)
  2. \(\frac{b}{h}\)
  3. \(\frac{p}{b}\)
  4. \(\frac{b}{p}\)
  5. \(\frac{h}{b}\)
  6. \(\frac{h}{p}\)

Above ratios are called trigonometric ratios and they are given certain names .Detail information is given below ;

Here,

Let Δ MNO be right angled triangle where ∠MNO = 90 ,

∠MNO = θ be the reference angle .

Then,

Side MN = Perpendicular ( p )

Side NO = Base ( b )

Side MO= Hypotenuse ( h )

Now , we can get six ratios here ;

  1. \(\frac{p}{h}\) = \(\frac{MN}{MO}\)
  2. \(\frac{b}{h}\) = \(\frac{NO}{MO}\)
  3. \(\frac{p}{b}\) = \(\frac{MN}{NO}\)
  4. \(\frac{b}{p}\) = \(\frac{NO}{MN}\)
  5. \(\frac{h}{b}\) = \(\frac{MO}{NO}\)
  6. \(\frac{h}{p}\) = \(\frac{MO}{MN}\)

Now , lets introduce the names for these ratios.

S.No. Ratio Nomenclature Abbreviation
1. \(\frac{p}{h}\) sine sin
2. \(\frac{b}{h}\) cosine cos
3. \(\frac{p}{b}\) tangent tan
4. \(\frac{b}{p}\) cotangent cot
5. \(\frac{h}{b}\) secant sec
6. \(\frac{h}{p}\) cosecant cosec/csc

Again , In the earlier statement if we try to link these ratio we get,

Sinθ = \(\frac{p}{h}\) =\(\frac{MN}{MO}\)

cosθ = \(\frac{b}{p}\) =\(\frac{NO}{MO}\)

tanθ = \(\frac{p}{b}\) =\(\frac{MN}{NO}\)

Cotθ =\(\frac{b}{p} \) =\(\frac{NO}{MN}\)

Secθ = \(\frac{h}{b}\) =\(\frac{MO}{NO}\)

Cosecθ=\(\frac{h}{p}\) = \(\frac{MO}{MN}\)

Operation Of Trignometric Ratios

Here, in trignometry ratio also, addition , subtraction , multiplication and division take place in sameway as in algebra .It is important to have the prior knowledge on algebra before operating trignometric ratios.

Addition

In algebra we can add terms just adding their coefficient .

for eg;

1. a + a = 2a

sin θ+ sinθ = 2sinθ

2. a + b = a + b

sinθ + cosθ = sinθ + cosθ

3. a2+ a = a2 + a

sec2 θ+ secθ = sec2θ + secθ

Subtraction

We can subtract the trigonometric ratios as we subtract in algebra.

1. 2a - a = a

2tanθ - tanθ = tanθ

2. a - b = a - b

cosecθ - cotθ = cosecθ - cotθ

3. a2- a = a2- a

sec2θ - secθ = sec2θ - secθ

Multiplication and Division

For, multiplication trigonometric ratios will also follow the laws of indices. Let's review laws of indicesonce .

a) a2× a3 = a ×a ×a ×a ×a = 5

Generalizing this statement : ap×aq= ap+q

b) a6÷ a2 = \(\frac{a×a ×a×a×a×a}{a×a}\)

= a4

Generaliziting ,

ap÷ aq = apq

c) ( a2)3= a2×a2×a2

= a2+2+2

=a6

Similary,

sin θ× sinθ = sin2θ

2sin2θ ×3sin3θ = 6sin5θ [ ap × aq = ap+q ]

6cos4θ ÷ 2cos2θ =\(\frac{6cos^4θ}{2cos^2θ}\) [ ap÷ aq = apq ]

( cot2θ)3 = cotθ6θ [ (ap)q÷ aq= ap×q]

3cosecθ× 4secθ = 12cosecθ.secθ

Some basic algebraical formula

S.No. Formula Expanded form Factorized form
1. (a + b)2 a2+2ab+b2 (a+b) (a+b)
2. (a - b)2 a2-2ab+b2 (a-b) (a-b)
3. a2 -b2 (a+b) (a-b)
4. a2+b2

(a+b)2 - 2ab

(a-b)2 +2ab

5. (a + b )3

a3+3a2b+3ab2+b3

a3+b3+3ab(a+b)

(a+b)(a+b)(a+b)
6. ( a- b)3

a3-3a2b+3ab2-b3

a3-b3-3ab(a-b)

(a-b)(a-b)(a-b)
7. a3+ b3 (a+b)3-3ab(a+b) (a+b) (a2-ab+b2)
8. a3- b3 (a-b)3+3ab(a-b) (a-b)(a2+ab+b2)



  • Multiplication trigonometric ratios follow the laws of indices.
  • We can subtract the trigonometric ratios as we subtract in algebra.
  • 6 ratios each of them obtained from these 3 sides they are :

    1. \(\frac{p} {h}\)
    2. \(\frac{b}{h}\)
    3. \(\frac{p}{b}\)
    4. \(\frac{b}{p}\)
    5. \(\frac{h}{b}\)
    6. \(\frac{h}{p}\)
.

Very Short Questions

Solution ;

Here, In ΔABC, ∠ABC = 90 ∠BAC = θ be the referance angle .

Then , 

Side BC = perpendicular ( p )

Side AC =  hypotenuse ( h )

Side AB = base  (b )

Now, using trignometry ratios formula we get;

sinθ = \(\frac{p}{h}\)                                        coscθ = \(\frac{b}{h}\)

tanθ =  \(\frac{p}{b}\)                                       cotθ =  \(\frac{b}{p}\)

secθ = \(\frac{h}{b}\)                                    cosecθ = \(\frac{h}{p}\)

a ) 

Solution;

        sinθ + 2sinθ

     =sinθ ( 1 + 2 )

     = sinθ ( 3 )

      =3 sinθ

 

 

b ) 

Solution; 

          5tanθ - 2tanθ

        =tanθ ( 5 - 2 )

       =tanθ ( 3 )

       =3 tanθ

        

a )sin θ × sin 2θ

               = sin θ × sin 2θ                     (   a× a n= am + n  )

               =sin3 θ

 

 

b ) ( cosec2θ - sec2θ )   by ( cosecθ +  secθ )    

  = \(\frac{cosec^2θ-sec^2θ}{cosecθ+secθ}\)

  =  \(\frac{(cosecθ)-(secθ)^2}{cosecθ+secθ}\)

  =\(\frac{(cosecθ+secθ)(cosecθ-secθ)}{cosecθ+secθ}\)

  =cosecθ-secθ

 

The methods of providing a trigonometric identity are as follows:

a. Take the identity on the left-hand side(L.H.S) and show it equal  to right-hand side(R.H.S).

b. Take the identity on the R.H.S and show it equal to L.H.S. 

Solution

= cos θ ( cosθ + sinθ)  - sinθ ( cosθ + sinθ)

= cos 2 θ + cosθ. sinθ - sinθ . cosθ - sin2θ

= cos 2 θ + cosθ. sinθ - cosθ . sinθ - sin2θ

= cos2θ - sin2θ

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  • These _____ rations are called trigonometric ratios .

     three
     two 
    six
     five
  • The full form of ratio sin  is _____

     cos
    sine
     tan 
  •  The abbreviated form of cotangent is _____ 

     tan
     cos
     sin
     cot
  • The abbreviated form of tangent is  _____ .

     cos
    tan
     sin
     cot
  • The ratio of sinθ = _____ .

    (frac{h}{b})
    (frac{p}{h})
    (frac{p}{b})
    (frac{b}{h})
  • The ratio of cos θ = _____  .

    (frac{h}{p})
    (frac{p}{h})
    (frac{b}{p})
    (frac{b}{h})
  • The ratio of tanθ = ______  

    (frac{h}{b})
    (frac{p}{b})
    (frac{h}{p})
    (frac{p}{h})
  • The abbreviated form of secant is _____ .

     cot
    sin
    tan
    sec 
  • The abbreviated form of cosecant____ . 

     sec 
     sin
    cosec
     cos
  • The full form of the ratio cos is_____ .

    cosine
    secant
     sine 
     cot
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