Question

if tan^2A + cot^2A = 10/3 then prove that tanA=√3 {please give an simple answer }

Answer 1

Answer:

tanA = √3

Step-by-step explanation:

Given,

tan^2 A + cot^2 A = 10 / 3

From the properties of trigonometry :

• tanB = 1 / cotB , B ≠ ( 2n + 1 )π / 2 for all n € N

Assuming that A ≠ ( 2n + 1 )π / 2, let the value of tan^2 A be x :

= > x + cot^2 A = 10 / 3

= > x + 1 / x = 10 / 3 { cot^2 A = 1 / tan^2 A = 1 / x }

= > ( x^2 + 1 ) / x = 10 / 3

= > ( x^2 + 1 ) / x = ( 9 + 1 ) / 3

= > ( x^2 + 1 ) / x = ( 3^2 + 1 ) / 3

On comparing both sides : x = 3

Thus,

= > tan^2 A = 3

= > tanA = √3

Hence, proved.

Answer 2

Question :-- if tan²A + cot²A = 10/3 , prove that TanA = √3.

Solution :---

→ tan²A + cot²A = 10/3

put cotA = 1/tanA

→ tan²A + 1/tan²A = 10/3

Now, put tan²A = x

→ x + 1/x = 10/3

Taking lcm ,

→ (x²+1)/x = 10/3

cross - Multiply now,

→ 3(x²+1) = 10x

→ 3x² - 10x + 3 = 0

Solving the Equation by splitting the middle term now,

→ 3x² - 9x - x +3 = 0

→ 3x(x-3) -1(x-3) = 0

→ (3x-1)(x-3) = 0

Puttiñg both Equal to zero now, we get,

→ 3x -1 = 0. Or, x - 3 = 0

→ 3x = 1. or, x = 3

→ x = 1/3

now, putting back x = tan²A , we get,

→ x = 3 = tan²A

→ tan²A = 3

Square root both sides

→ tanA = √3 . ( ✪✪ Hence Proved ✪✪)