Algebraic identities is the general term for equations that hold for all permissible values of their variables.
Solved Examples
Problem 1: Find the product of (x + 1)2 using standard algebraic identities.
Solution:
Given equation is (x + 1)2
Using identity (a+b)2 = a2 + 2ab + b2
put a = x and b = 1
(x+1)2 = x2 + 2.x.1 + 12
(x+1)2 = x2 +2x +1
Problem 2: Factorise (x4 – 1) using standard algebraic identities.
Solution:
Given equation (x4 – 1)
(x4 – 1) = [ (x2)2 - 1]
Using identity: (a+b)(a–b) = a2 – b2
(x4 – 1) = (x2+1)(x2-1)
Now again, (x2– 1) = (x+1)(x-1)
Hence, (x4 – 1) = (x2+1) (x+1)(x-1)
Problem 3: Expand (3x – 4y)2 using identity.
Solution:
Given (3x – 4y)2
a = 3x and b = 4y
Using (a–b)2 = a2 – 2ab + b2
(3x – 4y)2 = (3x)2 - 2.3x.4y + (4y)2
(3x – 4y)2 = 9x2 - 24xy + 16y2
Problem 4: Solve using identities: 122 - 102
Solution:
Given 122 - 102
Using (a+b)(a–b) = a2–b2
a = 12 and b = 10
Now, 122 - 102 = (12+10)(12-10)
122 - 102 = 22 ×2 = 44
Problem 5: Expand (2x + 3y + z)2 using identities.
Solution:
Given (2x+3y+z)2
Using: (a+b+c)2 = a2+b2+c2+2ab+2bc+2ac
a = 2x, b = 3y and c = z
(2x+3y+z)2 = (2x)2+(3y)2+(z)2+2.2x.3y+2.3y.z+2.z.2x
(2x+3y+z)2 = 4x2 + 9y2 + z2 + 12xy + 6yz + 4zx
Problem 6: If sinθ = 3/5, find cosθ.
Solution:
Given, sinθ = 3/5
Using Identity: sin2θ + cos2θ = 1
sin2θ = (3/5)2 = 9/25
cos2θ = 1 - sin2θ
= 1 - 9/25 = 16/25
cos2θ = 16/25
cosθ = 4/5
Problem 7: If tanθ = 3/4, find cosθ.
Solution:
Given, tanθ = 3/4
Using Identity: 1 + tan2θ = sec2θ
tan2θ = (3/4)2 = 9/16
sec2θ = 1 + tan2θ
= 1 + 9/16 = 25/16
sec2θ = 25/16
secθ = 5/4
Practice Questions
Q1. Expand (3x – 4y)3 using standard algebraic identities.
Q2. Use the identities to calculate the value of (81)2.
Q3. Use the identities to calculate the value of (37)2.
Q4. Use the identities to calculate the value of 97 × 103.
Q5. Simplify (2z + 5)2 – (2z - 5)2.
Q6. Use the identities to calculate the value of 96 × 102.
Q7. Use the identities to calculate the value of 91 × 109.
Q8. Use the identities to calculate the value of (87)2.
