Notes for introduction of trigonometry chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.
(1) For a right angled triangle ABC, side BC is called the side opposite to angle A, AC is called the hypotenuse and AB is called the side adjacent to angle A. (2) The trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides. The trigonometric ratios of angle A in right triangle ABC are defined as follows:
(i) sine of âˆ A = (side opposite to angle A)/ (hypotenuse) = BC/AC
(ii) cosine of âˆ A = (side adjacent to angle A)/ (hypotenuse) = AB/AC
(iii) tangent of âˆ A = (side opposite to angle A)/ (side adjacent to angle A) = BC/AB
(iv) cosecant of âˆ A = 1/( sine of âˆ A) = (hypotenuse)/ (side opposite to angle A) = AC/BC
(v) secant of âˆ A = 1/( cosine of âˆ A) = (hypotenuse)/ (side adjacent to angle A) = AC/AB
(vi) cotangent of âˆ A = 1/( tangent of âˆ A) = (side adjacent to angle A)/ (side opposite to angle A) = AB/BC
(3) Trigonometric Ratios of 45Â°:In Î” ABC, rightangled at B, if one angle is 45Â°, then the other angle is also 45Â°, i.e., âˆ A = âˆ C = 45Â°.
Now, let BC = AB = a.
Applying the Pythagoras Theorem, we get,
AC^{2} = AB^{2} + BC^{2} = a^{2} + a^{2 }= 2a^{2},
So, AC = a
As per the trigonometric ratio definitions, we have,
(i) sin 45Â° = (side opposite to angle 45Â°)/(hypotenuse) =BC/AC = Â a/ a = 1/
(ii) cos 45Â° = (side adjacent to angle 45Â°)/(hypotenuse) = AB/AC = a/ a = 1/
(iii) tan 45Â° = (side opposite to angle 45Â°)/(side adjacent to angle 45Â°) =BC/AB = a/ a = 1
(iv) cosec 45Â° = 1/( sin 45Â°) = (hypotenuse)/ (side opposite to angle 45Â°) = AC/BC =
(v) sec 45Â° = 1/( cos 45Â°) = (hypotenuse)/ (side adjacent to angle 45Â°) = AC/AB =
(vi) cot 45Â° = 1/( tan 45Â°) = (side adjacent to angle 45Â°)/ (side opposite to angle 45Â°) = AB/BC = 1.
(4) Trigonometric Ratios of 30Â° and 60Â°:Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60Â°, therefore, âˆ A = âˆ B = âˆ C = 60Â°. And draw a perpendicular AD from A to side BC.
Here, Î” ABD â‰… Î” ACD. Therefore, BD = DC and âˆ BAD = âˆ CAD (as per CPCT)
From the figure, it can be seen that, âˆ ABD = 60Â° and âˆ BAD = Â½ âˆ BAC = Â½ x 60Â° = 30Â°
Now, let us assume AB= BC = CA = 2a. Therefore, BD = Â½ BC = a.
Applying Pythagoras theorem in Î” ABD, we get,
AD^{2} = AB^{2} â€“ BD^{2} = (2a)^{2} â€“ (a)^{2} = 3a^{2
}Hence, AD = a
As per the trigonometric ratio definitions, we have,
(i) sin 30Â° = BD/AB = a/2a = Â½
(ii) cos 30Â° = AD/AB = a /2a = /2
(iii) tan 30Â° = BD/AD = a/a = 1/
(iv) cosec 30Â° = 1/ sin 30Â° = 2
(v) sec 30Â° = 1/ cos 30Â° = 2/
(vi) cot 30Â° = 1/ tan 30Â° =
Similarly, as per the trigonometric ratio definitions, we have,
(i) sin 60Â° =/2
(ii) cos 60Â° = 1/2
(iii) tan 60Â° =
(iv) cosec 60Â° = 2
(v) sec 60Â° = 2
(vi) cot 60Â° = 1/
(5) Trigonometric Ratios of 0Â°:
(i) sin 0Â° = 0
(ii) cos 0Â° = 1
(iii) tan 0Â° = 0
(iv) cosec 0Â° = Not defined
(v) sec 0Â° = 1
(vi) cot 0Â° = Not defined
(6) Trigonometric Ratios of 90Â°:
(i) sin 90Â° = 1
(ii) cos 90Â° = 0
(iii) tan 90Â° = Not defined
(iv) cosec 90Â° = 1
(v) sec 90Â° = Not defined
(vi)cot 90Â° = 0
(7) Table representing the trigonometric ratios of 0Â°, 30Â°, 45Â°, 60Â° and 90Â°:
âˆ A 
0Â°  30Â°  45Â°  60Â°  90Â° 
sin A 
0 
1/2 


1 
cos A  1  1/2 
0 

tan A 
0 

1 
Â 
Not defined 
cosec A  Not defined  2  Â 
1 

sec A 
1 

Â 
2 
Not defined 
cot a  Not defined  Â  1 
0 
For Example:Â Evaluate 2 tan^{2} 45Â° + cos^{2} 30Â° â€“ sin^{2} 60Â°
Now, 2 tan^{2} 45Â° + cos^{2} 30Â° â€“ sin^{2} 60Â°
= 2 (1)^{2} + ()^{2} â€“ ()^{2
}= 2 + Â¾  Â¾
= 2.
(8) Trigonometric Ratios of Complementary Angles:
(i) sin (90Â° â€“ A) = cos A
(ii) cos (90Â° â€“ A) = sin A
(iii) tan (90Â° â€“ A) = cot A
(iv) cot (90Â° â€“ A) = tan A
(v) sec (90Â° â€“ A) = cosec A
(vi) cosec (90Â° â€“ A) = sec A
For Example:Â Simplify tan 26Â° / cot 64Â°.
(i) We know that, cot A = tan (90Â° â€“ 64Â°) = cot 64Â°.
(ii) Therefore, tan 26Â° / cot 64Â° = cot 64Â°/ cot 64Â° = 1.
(9) Trigonometric Identities:
(i) cos^{2} A + sin^{2} A = 1
(ii) 1 + tan^{2} A = sec^{2} A
(iii) cot^{2} A + 1 = cosec^{2} A
For Example:Â Evaluate sin 25Â° cos 65Â° + cos 25Â° sin 65Â°.
=sin 25Â° cos 65Â° + cos 25Â° sin 65Â°
= sin 25Â° cos (90Â°  25Â°) + cos 25Â° sin( 90Â°  25Â°)
= sin^{2} 25Â° + cos^{2} 25Â°
= 1 (Since cos^{2} A + sin^{2} A = 1)
For Example:Â Prove that (cosec A â€“ sin A)(sec A â€“ cos A) = 1/(tan A + cot A)
LHS = (cosec A â€“ sin A)(sec A â€“ cos A)
= (1/sin A â€“ sin A)(1/cos A â€“ cos A)
= ((1â€“ sin^{2} A)/sin A) ((1â€“ cos^{2} A)/cos A)
=( cos^{2} A x sin^{2} A)/(sin A cos A)
= sin A cos A
RHS = 1/(tan A + cot A)
= 1/(sin A/cos A + cos A/sin A)
= 1/((sin^{2} A + cos^{2} A)/(sin A cos A))
= (sin A cos A)/( sin^{2} A + cos^{2} A)
= sin A cos A
Thus, LHS = RHS.Â
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