Subject: Optional Mathematics
Let us consider a right-angled triangle ABC in which ∠ABC = 90°. This triangle consists of a right angle, two acute angles and three sides. These are called elements of a right-angled triangle. 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. This side opposite to the right angle is called hypotenuse and it is denoted b
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.
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
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.
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 = \(\frac{x}{y}\)
There are generally three fundamental relations of trigonometric ratios.
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 = \(\frac{x}{y}\)
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 = \(\frac{1}{Tanθ}\) |
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 = \(\frac{cosecθ}{secθ}\) |
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)
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.
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 = \(\frac{b}{p}\) = \(\sqrt{cosec^2θ-1}\)
Angles→↓ | 0° | 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 :
Multiply:
(sinA + sinB) (sinA - sinB)
Soln:
Here, (sinA + sinB) (sinA - sinB)
= sinA (sinA - sinB) + sinB (sinA - sinB)
= sin2A - sinA.sinB + sinA.sinB - sin2B
= sin2A - sin2B
Multiply:
(1 - cosθ) ( 1 + cosθ)
Soln:
Here, ( 1 - cosθ) ( 1 + cosθ)
= 1 - cos2θ
= sin2θ. Ans
Multiply:
(1 + tanθ) (1 - tanθ) (1 + tan2θ)
Soln:
Here, ( 1+ tanθ) ( 1 - tanθ) ( 1 + tan2θ)
= (1 - tan2θ) (1 + tan2θ) = 1 - tan4θ Ans.
Factorize :
cos2A - sin2A
Soln:
cos2A - sin2A = ( cosA - sinA) (cosA + sinA). Ans.
Prove the following:
(1 + tan2A) cos2A = 1
Soln:
L. H. S. = (1 + tan2A) cos2A
= sec2A. cos2A
= \(\frac{1}{cos^2A}\). cos2A
= 1 = RHS Proved.
Prove the following:
\(\frac{1}{cos^2A}\) - \(\frac{1}{cot^2A}\) = 1
Soln:
LHS =\(\frac{1}{cos^2A}\) - \(\frac{1}{cot^2A}\) = sec2A -- tan2A = 1 + tan2A - tan2A
=1 = RHS proved
Prove the following:
\(\frac{secA}{cosA}\) - \(\frac{tanA}{cotA}\) = 1
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
Prove that:
\(\frac{1-tanA}{1+tanA}\) =\(\frac{cotA - 1}{cotA + 1}\)
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
Prove that:
(tanθ + secθ)2 = \(\frac{1 + sinθ}{1 - sinθ}\)
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.
Prove that:
\(\frac{1}{1 - cosA}\) - \(\frac{1}{1 + cosA}\) = 2cotA cosecA
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.
Prove that:
\(\frac{cosx}{1 - sinx}\) + \(\frac{cosx}{1 + sinx)}\) = 2secx
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
Prove that:
sin4θ + cos4θ = 1 - 2sin2θ cos2θ
Soln:
Taking LHS,
=sin4θ + cos4θ = (sin2θ)2 + (cos2θ)2
=(sin2θ + cos2θ)2 - 2sin2θ cos2θ
= 1 - 2sin2θ cos2θ = RHS Proved
Prove that:
(1 + sinA + cosA)2 = 2(1 + sinA) (1 + cosA)
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
Prove that:
\(\frac{cosA - sinA + 1}{cosA + sinA - 1}\) =\(\frac{1 + cosA}{sinA}\)
Soln:
LHS =\(\frac{cosA - sinA + 1}{cosA + sinA - 1}\)
=\(\frac{\frac{cosA - sinA + 1}{sinA}}{\frac{cosA + sinA - 1}{sinA}}\)
=\(\frac{cotA - 1 + cosecA}{cotA + 1 - cosecA}\)
= \(\frac{(cotA + cosecA) - (cosec^2 A - cot^2 A)}{cotA - cosecA + 1}\)
=\(\frac{1 (cotA + cosecA) - (cosecA - cotA) (cosecA + cotA) }{cotA - cosecA + 1}\)
= \(\frac{(cosecA + cotA) (1 - cosecA + cotA)}{cotA - cosecA + 1}\)
= cosecA + cotA =\(\frac{1}{sinA}\) + \(\frac{cosA}{sinA}\)
= \(\frac{1 + cosA}{sinA }\) =RHS proved.
Prove that:
\(\frac{secA - tanA}{secA + tanA}\) = 1 - 2 secA tanA + 2 tan2A.
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.
Prove that:
(secA + cosecA)2 = ( 1+ cotA)2 + (1 + tanA)2
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
sec4θ - cosec4θ
soln: Here, sec4θ - cosec4θ
= (sec2θ)2 - (cosec2θ)2
= ( sec2θ - cosec2θ) (sec2θ + cosec2θ)
= ( secθ - cosecθ ) (secθ - cosecθ ) (sec2θ + cosec2θ)
sin2x + 3 sinx + 2
soln: here, sin2x + 3sinx + 2
= sin2x + 2sinx + sinx + 2
= sinx (sinx + 2) +1 (sinx + 2)
= (sinx + 2 ) (sinx + 1)
(1 - cos2 A) (1 + cot2 A)
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
tanθ . \(\sqrt{1}-{sin^2θ}\) = sinθ
soln: L.H.S=tanθ . \(\sqrt{1} {sin^2θ}\) = tanθ \(\sqrt{cos^2θ}\)
= tanθ . cosθ = \(\frac{sinθ}{cosθ}\).cosθ=sinθ= R.H.S. proved
tan2 A - sin2 A = sin2 A . tan2 A
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.
\(\frac{1}{secA-tanA}\)-\(\frac{1}{cosA}\)= \(\frac{1}{cosA}\)- \(\frac{1}{secA+tanA}\)
soln; L.H.S=\(\frac{1}{secA-tanA}\)-\(\frac{1}{cosA}\)
= \(\frac{sec^2A-tan^2A}{secA-tanA}\)-\(\frac{1}{cosA}\)
= \(\frac{(secA+tanA)(secA-tanA}{(secA-tanA}\)-secA
= secA - secA + tanA = tanA
∴ L.H.S.= R.H.S. proved.
R.H.S = \(\frac{1}{cosA}\)-\(\frac{1}{secA + tanA}\)
= secA-\(\frac{sec^2+tan^2}{secA + tanA}\)
= sec A-\(\frac{(secA+tanA)(secA-tanA)}{(secA+tanA)}\)
= secA - secA + tanA=tanA
∴ L.H.S=R.H.S. proved.
cosec4A (1-cos4A) = 1+2 cot2A
soln:L.H.S. = cosec4A (1-cos4A)
= cosec4A(1-cos2 A) (1+cos2 A)= cosec4θ[1-(cos2A)2]
= cosec2 A. cosec2 A. sin2 A (1+cos2A)
= cosec2A.\(\frac{1}{sin^2A}\).sin2A(1+cos2A)
= cosec2A (1+cos2A)
= cosec2A+cosec2A cos2A= cosec2A + \(\frac{cos^2A}{sin^2A}\)
= cosec2A + cot2 A = 1 + cot2A+cot2A
= 1+2 cot2A= R.H.S. proved.
sin3θ= 3 sinθ - 4 sin3θ
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.
sin (∝+ß) = sin∝cosß + cos ∝ sinß
soln: given,∝=00,β=300
L.H.S.= sin(∝ + β) = sin (00 + 300)= sin300=\(\frac{1}{2}\)
R.H.S = sin∝ cosβ + cos∝ sinβ
= sin00 cos00 + cos00 sin00
= 0×\(\frac{\sqrt{3}}{2}\)+ 1 \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\) = 0 +\(\frac{1}{2}\)=\(\frac{1}{2}\)
∴ L.H.S = R.H.S. proved.
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