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Note on Trigonometrical Ratio

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Angle

One of the acute angles is considered as the reference angle. This reference angle is used for naming the sides of a right-angled triangle. The side opposite to the reference angle is called perpendicular and it is denoted by the letter p. The side opposite to the right angle is called hypotenuse and it is denoted by the letter h. The remaining side is base and it is denoted by b. Clearly, the base is the side between right angle and reference angle.Let us consider a right-angled triangle ABC in which ∠ABC = 90°. This triangle consists of a right angle, two acute and three sides. There are called elements of a right-angled triangle.

If ∠ABC = 0 is taken as reference angle, then AB, AC and BC are perpendiculars, hypotenuse and base of the right-angled triangle ABC. If ∠BAC =Φ is the reference angle, then BC, AC and AB respectively are perpendicular, hypotenuse and base of the same right-angled triangle ABC. So, the name of the sides of a right-angled triangle depends on the choice of the reference angle.

With the help of three sides of a right-angled triangle, six ratios can be derived taking anyone acute angle as the reference angle.

Six Trigonometric Ratios

i) \(\frac{AB}{BC}\)

ii) \(\frac{AB}{CA}\)

iii) \(\frac{BC}{AB}\)

iv) \(\frac{BC}{CA}\)

v) \(\frac{CA}{AB}\)

vi) \(\frac{CA}{BC}\)

These ratios are commonly known as trigonometric ratios.

If \(\angle\)ABC = 0 is taken as reference angle, then

AB = perpendicular(p), CA = hypotenuse(h) and, BC = base(b)

Now, the six trigonometric ratios of ΔABC right angled at B with reference angle Θ

  1. \(\frac{AB}{BC}\) = \(\frac{p}{b}\) = Tangent of angle θ = Tanθ
  2. \(\frac{AB}{CA}\) = \(\frac{p}{h}\) = Sine of angle θ= Sinθ
  3. \(\frac{BC}{AB}\) = \(\frac{b}{p}\) = Cotangent of angle θ = Cotθ
  4. \(\frac{BC}{CA}\) = \(\frac{b}{h}\) = Cosine of angle θ = Cosθ
  5. \(\frac{CA}{AB}\) = \(\frac{h}{p}\) = Cosecant of angle θ = Cosecθ
  6. \(\frac{CA}{BC}\) = \(\frac{h}{b}\) = Secant of angle θ = Secθ

Traditionally, trigonometric ratios are defined on the angles of a triangle as above. Now a days these trigonometric ratio are defined on the angles of any magnitude as below.

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Let us consider an angle θ placed in the standard position. Draw circle with centre at 0 and radius r. Let the circle intersect the terminal line at some point P(x,y) as shown in the figure. Draw perpendicular PM from P to the x-axis. Then, PM = y, OM = x and OP = r.

For any angle θ, the six trigonometric ratios can be defined in terms of x-coordinate 'x', y-coordinate 'y' can the radius 'r' by the formula.

Sinθ = \(\frac{y}{r}\),  Cosθ = \(\frac{x}{r}\),  Tanθ = \(\frac{y}{x}\),  Cosecθ = \(\frac{r}{y}\),  Secθ = \(\frac{r}{x}\),  Cotθ = \(\frac{x}{y}\)

Fundamental relations of trigonometric ratios

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There are generally three fundamental relations of trigonometric ratios.

  1. Reciprocal Relation
  2. Quotient Relation
  3. Pythagorean Relation

Let us consider a circle centered at the origin O and having radius r. Let P(x,y) be any point on the circle. Join OP and let ∠XOP = θ. Draw perpendicular PM from P to X-axis. Then OP = r, PM =y and OM = x.

Sinθ = \(\frac{y}{r}\), Cosθ = \(\frac{x}{r}\), Tanθ = \(\frac{y}{x}\), Cosecθ = \(\frac{r}{y}\), Secθ = \(\frac{r}{x}\), Cotθ = \(\frac{x}{y}\)

Reciprocal Relations

According to trignometric ratios

sinθ × cosecθ = \(\frac{y}{r}\) × \(\frac{r}{y}\) = 1 

∴ sinθ = \(\frac{1}{cosecθ}\) ............(i) 

∴ cosecθ = \(\frac{1}{sinθ}\) ............(ii)

Similarly, cosθ × secθ = \(\frac{x}{r}\) × \(\frac{r}{x}\) = 1 

∴ cosθ = \(\frac{1}{secθ}\) ............(iii) 

∴ secθ = \(\frac{1}{cosθ}\) ............(iiv)

Similarly, tanθ × cotθ = \(\frac{x}{r}\) × \(\frac{r}{x}\) = 1 

∴ tanθ = \(\frac{1}{cotθ}\) ............(v) 

∴ cotθ = \(\frac{1}{tanθ}\) ............(vi)

 

Hence the reciprocal relations are:

Sinθ·Cosecθ = 1 Sinθ = \(\frac{1}{Cosecθ}\) Cosecθ = \(\frac{1}{Sinθ}\)
Cosθ·Secθ =1 Cosθ = \(\frac{1}{Secθ}\) Secθ = \(\frac{1}{Cosθ}\)
Tanθ·Cotθ =1 Tanθ =\(\frac{1}{Cotθ}\) Cotθ = \(\frac{1}{Tanθ}\)
 
Quotient Relations

We have, Tanθ= \(\frac{y}{x}\) = \(\frac{\frac{y}{r}}{\frac{x}{r}}\) = \(\frac{sinθ}{cosθ}\) = \(\frac{secθ}{cosecθ}\)

Again, Cotθ =\(\frac{x}{y}\) = \(\frac{\frac{x}{r}}{\frac{y}{r}}\) = \(\frac{cosθ}{sinθ}\) = \(\frac{cosecθ}{sinθ}\)

Hence, the quotient relations are:

tanθ = \(\frac{sinθ}{cosθ}\) cotθ = \(\frac{cosθ}{sinθ}\)
tanθ = \(\frac{secθ}{cosecθ}\) cotθ = \(\frac{cosecθ}{secθ}\)
 
Pythagorean Relations

By Pythagoras theorem,

In right angled ΔPOM, x2 + y2 = r2

Dividing both sides by r2 we get

\(\frac{x^2}{r^2}\) + \(\frac{y^2}{r^2}\) = 1

i.e., sin2θ + cos2θ= 1............(i)

From equation (i) 

∴ sin2θ = 1 - cos2θ............(ii)

and cos2 = 1 - sin2θ..........(iii)

Now, Dividing both sides by sin2θ in (i), we get

\(\frac{sin^2θ}{sin^2θ}\) + \(\frac{cos^2θ}{sin^2θ}\) = \(\frac{1}{sin^2θ}\)

or, 1+ cot2θ = cosec2θ

∴ cosec2θ - cot2θ=1 .........(iv) 

Again, Dividing both sides by cos2θ in (i), we get

\(\frac{sin^2θ}{cos^2θ}\) + \(\frac{cos^2θ}{cos^2θ}\) = \(\frac{1}{cos^2θ}\)

or, tan2θ + 1 = sec2θ

∴ sec2θ - tan2θ =1 .........(v)

Conversion of Trigonometric Ratios

One trigonometrical ratio can be expressed as other trigonometrical ratios. sinθ can be expressed into cosecθ, secθ tanθ cotθ and cosθ by using trigonometrical formulae. Similarly, cosecθ, secθ, tanθ, cotθ, and cosθ can be expressed into sinθ. There are two methods for the conversion of trigonometrical ratios.

  1. Direct or formula method
  2. Indirect or geometrical method.

Direct or formula method

Express all the trigonometrical ratio in terms of cosθ by using direct method.

Solution:

Using trigonometrical formula to convert in terms of cosθ, 

sinθ = \(\sqrt{1-cos^2θ}\)

cosθ = cosθ

cosecθ = \(\frac{1}{sinθ}\) = \(\frac{1}{1-cos^2θ}\)

tanθ = \(\frac{sinθ}{cosθ}\) = \(\frac{1-cos^2θ}{cosθ}\)

secθ = \(\frac{1}{cosθ}\)

cotθ = \(\frac{cosθ}{sinθ}\) = \(\frac{cosθ}{1-cos^2θ}\)

Indirect or geometrical method

Express all the trigonometrical ratios in terms of cosecθ,

Solution,

ΔABC is a right angle triangle where ∠B=90o

Then,

sinθ = \(\frac{p}{h}\) = \(\frac{1}{k}\) = \(\frac{1}{cosecθ}\)

cosθ =\(\frac{b}{h}\) =\(\frac{cosec^2θ-1}{cosecθ}\)

tanθ =\(\frac{p}{b}\) = \(\frac{1}{cosec^2θ-1}\)

cosecθ = cosecθ

secθ = \(\frac{h}{b}\) = \(\frac{cosecθ}{cosec^2θ-1}\)

cotθ = \(\frac{b}{p}\) = \(\sqrt{cosec^2θ-1}\)

Value of Trigonmetric Ratios

Angles 30° 45° 60° 90°
sinθ 0 \(\frac{1}{2}\) \(\frac{1}{\sqrt2}\) \(\frac{\sqrt{3}}{2}\) 1
cosθ 1 \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt2}\) \(\frac{1}{2}\) 0
tanθ 0 \(\frac{1}{\sqrt3}\) 1 \(\sqrt{3}\)
cosecθ 2 \(\sqrt{2}\) \(\frac{2}{\sqrt3}\) 1
secθ 1 \(\frac{2}{\sqrt3}\) \(\sqrt{2}\) 2
cotθ \(\sqrt{3}\) 1 \(\frac{1}{\sqrt3}\) 0

The six trigonometric ratio of ΔABC right angled at B with reference angle Θ are as follow :

  • \(\frac{AB}{BC}\) = \(\frac{p}{b}\) = Tangent of angle θ = Tanθ
  • \(\frac{AB}{CA}\) = \(\frac{p}{h}\) = Sine of angle θ= Sinθ
  • \(\frac{BC}{AB}\) = \(\frac{b}{p}\) = Cotangent of angle θ = Cotθ
  • \(\frac{BC}{CA}\) = \(\frac{b}{h}\) = Cosine of angle θ = Cosθ
  • \(\frac{CA}{AB}\) = \(\frac{h}{p}\) = Cosecant of angle θ = Cosecθ
  • \(\frac{CA}{BC}\) = \(\frac{h}{b}\) = Secant of angle θ = Secθ
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Very Short Questions

Soln:

Here, (sinA + sinB) (sinA - sinB)

= sinA (sinA - sinB) + sinB (sinA - sinB)

= sin2A - sinA.sinB + sinA.sinB - sin2B

= sin2A - sin2B

Soln:

Here, ( 1 - cosθ) ( 1 + cosθ)

= 1 - cos2θ

= sin2θ. Ans

Soln:

Here, ( 1+ tanθ) ( 1 - tanθ) ( 1 + tan2θ)

= (1 - tan2θ) (1 + tan2θ) = 1 - tan4θ Ans.

Soln:

cos2A - sin2A = ( cosA - sinA) (cosA + sinA). Ans.

Soln:

L. H. S. = (1 + tan2A) cos2A

= sec2A. cos2A

= \(\frac{1}{cos^2A}\). cos2A

= 1 = RHS Proved.

Soln:

LHS =\(\frac{1}{cos^2A}\) - \(\frac{1}{cot^2A}\) = sec2A -- tan2A = 1 + tan2A - tan2A

=1 = RHS proved

Soln:

L.H.S. =\(\frac{secA}{cosA}\) - \(\frac{tanA}{cotA}\) = secA. \(\frac{1}{cosA}\) - tanA. \(\frac{1}{cotA}\)

= secA. secA - tanA. tanA

= sec2A - tan2A = 1 = R. H. S. Proved

Soln:

Here,\(\frac{1-tanA}{1+tanA}\) =\(\frac{\frac{1}{tanA}-\frac{tanA}{taanA}}{\frac{1}{tanA}+\frac{tanA}{tanA}}\)

= \(\frac{cotA - 1}{cotA + 1}\) RHS proved

Soln:

L.H.S = ( tanθ + secθ)2

= (\(\frac{sinθ}{cosθ}\) + \(\frac{1}{cosθ}\))2

= (\(\frac{1 + sinθ}{cosθ}\))2

= \(\frac{(1 + sinθ)( 1 + sinθ)}{cos^2θ}\)

=\(\frac{(1 + sinθ)(1 + sinθ)}{1 - sin^2θ}\)

= \(\frac{1 + sinθ}{1 - sinθ}\) = RHS Proved.

Soln:

LHS. =\(\frac{1}{1 - cosA}\) - \(\frac{1}{1 + cosA}\) = \(\frac{1 + cosA -(1 - cosA)}{(1 - cosA) ( 1 + cosA)}\)

= \(\frac{1 + cosa - 1 + cosA}{1 - cos^2 A}\)

= \(\frac{2cosA}{sin^2 A}\) =\(\frac{2cosA}{sinA}\) .\(\frac{1}{sinA}\)

= 2cotA cosecA = RHS Proved.

Soln:

Taking LHS,
=\(\frac{cosx}{1 - sinx}\) + \(\frac{cosx}{1 + sinx)}\)

= \(\frac{cosx(1 + sinx) + cosx(1 - sinx)}{(1 - sinx) (1 + sinx)}\)

= \(\frac{cosx + cosx.sinx + cosx - cosx.sinx}{1 - sin^2 x}\)

= \(\frac{2cosx}{cos^2x}\) =\(\frac{2}{cosx}\) = 2secx = RHS proved. Ans

Soln:
Taking LHS,
=sin4θ + cos4θ = (sin2θ)2 + (cos2θ)2

=(sin2θ + cos2θ)2 - 2sin2θ cos2θ

= 1 - 2sin2θ cos2θ = RHS Proved

Soln:

LHS = ( 1 + sinA + cosA)2

={(1 + sinA) +cosA}2

= (1 + sinA)2 + 2(1 + sinA) cosA + cos2A

=(1 + sinA)2 + 2( 1 + sinA) cosA + cos2A

= ( 1+ sinA)2+ 2(1 + sinA) cosA + 1 - sin2A

= ( 1+ sinA)2 + 2(1 + sinA) cosA + (1 + sinA) (1 - sinA)

=(1 + sinA) ( 1 + sinA + 2cosA + 1 -sinA)

= ( 1 + sinA) ( 2 + 2 cosA)

= 2(1 + sinA) (1 + cosA) = RHS Proved

Soln:

LHS =\(\frac{secA - tanA}{secA + tanA}\) =\(\frac{(secA - tanA)}{secA + tanA}\)× \(\frac{secA - tanA}{secA - tanA}\)

= \(\frac{(secA - tanA)^2}{sec^2 A -tan^2A}\)

=\(\frac{sec^2 A - 2secA tanA + tan^2A}{1}\)

= 1 + tan2A - 2secA tanA + tan2A

= 1 - 2secA tanA + 2 tan2A = RHS proved.

Soln:

RHS = 1 + 2cotA + cot2A + 1 + 2 tanA + tan2A

= 1 + cot2A + 1 + tan2A + 2 (cotA + tanA)2

= 1 + cot2A + 1 + tan2A + 2 (cotA + tanA)

= cosec2A + sec2A + 2(\(\frac{cosA}{sinA}\) + \(\frac{sinA}{cosA}\))

=coscec2A +sec2A + 2(\(\frac{cos^2A + sin^2A}{sinA cosA}\))

= cosec2A + sec2A + 2. \(\frac{1}{sinA cosA}\)

= cosec2A + sec2A + 2coosecA secA

= (secaA + cosecA)2 =LHS proved

soln: Here, sec4θ - cosec4θ

= (sec2θ)2 - (cosec2θ)2

= ( sec2θ - cosec2θ) (sec2θ + cosec2θ)

= ( secθ - cosecθ ) (secθ - cosecθ ) (sec2θ + cosec2θ)

soln: here, sin2x + 3sinx + 2

= sin2x + 2sinx + sinx + 2

= sinx (sinx + 2) +1 (sinx + 2)

= (sinx + 2 ) (sinx + 1)

soln: L.H.S =(1 - cos2 A ) (1 + cot2 A ) = 1 ∴ 1 + cot2θ

= sin2 A . cosec2 A = sin2 A× \(\frac{1}{sin^2}A\) = cosec2 θ

= 1 = R.H.S proved. and 1 - cos2θ = sin2θ

soln: L.H.S=tanθ . \(\sqrt{1} {sin^2θ}\) = tanθ \(\sqrt{cos^2θ}\)

= tanθ . cosθ = \(\frac{sinθ}{cosθ}\).cosθ=sinθ= R.H.S. proved

soln: L.H.S = tan2 A - sin2 A = \(\frac{sin^2A}{cos^2A}\) - sin2 A

= \(\frac{sin^2 A- sin^2 A cos^2 A}{cos^2}\)=\(\frac{sin^2 (1 - cos^2 A)}{cos^2}\)

= sin2A . \(\frac{sin^2A}{cos^2A}\)= sin2 A. tan2 A

= R.H.S. proved.

soln: Given,θ = 300

L.H.S. = sin3θ = sin 3× 300 = sin900 = 1

R.H.S. = 3sinθ - 4sin3θ = 3sin300- 4sin3300

= 3×\(\frac{1}{2}\)-4 \(\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}\)3 = \(\frac{3}{2}\)- \(\frac{4}{8}\)

= \(\frac{3}{2}\)- \(\frac{1}{2}\)= \(\frac{3-1}{2}\)=\(\frac{2}{2}\)=1

∴ L.H.S.= R.H.S. proved.

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Krischal Dhungel

if sinA = m/n then show that (n^2-m^2)^1/2 tanA = m


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