Introduction to Trigonometry
Mathematics Class Tenth
Trigonometric identities & NCERT Exercise 8.4
Trigonometric identities
An equation contains trigonometric ratios is called a Trigonometric Indentity, it is true for all values of the angle(s) involved.
Obtaining Trigonometric Identity
Let ABC is a right angle triangle, in which ∠C = 90o
In this right angle triangle,
Hypotenuse (p) = AB
And, for the acute angle A,
Base (b) = AC
And, Perpendicular (p) = BC
Now according to the Pythagoras Theorem, we know that
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
Thus, In the right angle triangle ABC
(AB)2 = (BC)2 + (AC)2 - - - - (i)
After dividing each term of the equation (i) by AB, we get
[AB/AB]2 = [BC/AB]2 + [AC/AB]2
⇒ 1 = [BC/AB]2 + [AC/AB]2
⇒ 1 = sin2 A + cos2 A
[∵ sin A = Perpendicular/Hypotenuse = BC/AB and cos A = Base/Hypotenuse = AC/AB]
⇒ sin2 A + cos2 A = 1 - - - - (ii)
⇒ sin2 A = 1 – cos2A - - - - (iii)
⇒ cos2 A = 1 – sin2A - - - - (iv)
Now, after dividing equation (i) by BC, we get
[AB/BC]2 = [BC/BC]2 + [AC/BC]2
[AB/BC]2 = 1 + [AC/BC]2
⇒ cosec2 A = 1 + cot2 A - - - - (v)
[∵ cosec A = Hypotenuse/Perpendicular = AB/BC, and cot A = Base/Perpendicular]
⇒ cosec2 A – cot2 A = 1 - - - - (vi)
⇒ cosec2 A – 1 = cot2 A - - - - (vii)
Now, after dividing equation (i) by AC, we get
[AB/AC]2 = [BC/AC]2 + [AC/AC]2
⇒ [AB/AC]2 = [BC/AC]2 + 1
⇒ sec2 A = tan2 A + 1
[∵ sec A = Hypotenuse/Base, and tan A = Perpendicular/Base]
⇒ 1 + tan2 A = sec2 A - - - - (viii)
⇒ tan2 A = sec2 A – 1 - - - - (ix)
⇒ sec2 A – tan2 A = 1 - - - - (x)
Solution of NCERT Exercise 8.4 question 1 and 2
Question (1) Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
Expression of sin A in terms of cot A
We know that, sin A = 1/cosec A
By squaring both sides, we get
(sin A)2 = (1/cosec A)2
(sin A)2 = 1/cosec2A
⇒ sin A = 1/cosec2A
sin A = 1/cosec2A
After substituting trigonometric identity cosec2A = 1 + cot2A, we get
sin A = 1/1 + cot2A
Expression of sec A in terms of cot A
We know that, sec2A = 1 + tan2A
⇒ sec2A = 1 + 1/cot2A
⇒ sec2A = cot2A + 1/cot2A
sec2A = cot2A + 1/cot2A
Expression of tan A in terms of cot A
We know that tan A = 1/cot A
Thus,
sin A = 1/1 + cot2A
sec2A = cot2A + 1/cot2A, and
tan A = 1/cot A
Question (2) Write all the other trigonometric ratios of ∠ A in terms of sec A.
Solution:
Expression of cos A in sec A
We know that
cos A = 1/sec A - - - - - - (i)
Expression of sin A in sec A
We know that
sin2 A = 1 – cos2A
= 1 – 1/sec2A
[∵ cos A = 1/secA]
= sec2A – 1/sec2A
∴ sin A = sec2A – 1/sec2A
⇒ sin A = sec2A – 1/sec A - - - - (ii)
Expression of cosec A in sec A
We know that, cosec A = 1/sin A
After substituting value of sin A from equation (ii), we get
cosec A = 1/sec2A – 1/sec A
⇒ cosec A = sec A/sec2 A – 1 - - - - (iii)
Expression of tan A in sec A
We know that, 1 + tan2A = sec2A
∴ tan2 A = sec2A – 1
⇒ tan A = sec2A – 1 --------- (iv)
Expression of cot A into sec A
We know that, cot A = 1/tan A
After substituting the value of tan A from equation (iv) we get
cot A = 1/sec2A – 1 --------(v)
Thus,
sin A = sec2A – 1/sec A
cos A = 1/sec A
tan A = sec2A –1
cosec A = sec A/sec2A – 1
And, cot A = 1/sec2A – 1 Answer
Reference: