Question

If 3 cot A = 4, Check whether `((1-tan^2 A)/(1+tan^2 A)) = cos^2 "A" - sin^2 "A"` or not.

Answer 1

It is given that 3cot A = 4 or cot A = `4/3`

Consider a right triangle ABC, right-angled at point B.

cot A =` ("Side adjacent to ∠A")/("Side opposite to ∠A")`

`("AB")/("BC") =  4/3`

If AB is 4k, then BC will be 3k, where k is a positive integer.

In ΔABC,

(AC)² = (AB)² + (BC)²

= (4k)² + (3k)²

= 16k² + 9k²

= 25k²

AC = 5k

cos A = `("Side adjacent to ∠A")/"Hypotenuse" = ("AB")/("AC")`

= `(4k)/(5k)`

= `4/5`

sin A = `("Side adjacent to ∠A")/"Hypotenuse" = ("BC")/("AC")`

= `(3k)/(5k)`

= `3/5`

tan A = `("Side adjacent to ∠A")/"Hypotenuse" = ("BC")/("AB")`

= `(3k)/(4k)`

= `3/4`

`(1-tan^2 A)/(1+tan^2 A) = ((1 - (3/4)^2)/(1+(3/4)^2))`

= `((1-9/16)/(1+9/16))`

= `(7/16)/(25/16)`

= `7/25`

`cos^2 A + sin^2 A = (4/5)^2 - (3/5)^2`

= `16/25 - 9/25`

= `7/25`

∴ `(1-tan^2A)/(1+tan^2 A)= cos^2A - sin^2A`

Answer 2

3 cot A = 4, check = `(1 - tan^2 A)/(1 + tan^2 A) = cos^2 A - sin^2 A`

cot A = `"adjacent side"/"opposite side" = 4/3`

Let x be the hypotenuse

By Applying Pythagoras theorem

AC² = AB² + BC²

x² = 4² + 3²

x² = 25²

x = 5

tan A = `1/(cos^2 A)` = `3/4`

cos A = `"adjacent side"/"hypotenuse" = 4/5`

sin A = `3/5`

L.H.S = `(1 - tan^2 A)/(1 + tan^2 A)`

= `(1 - (3/4)^2)/(1 + (3/4)^2)`

= `((16 - 9)/16)/((16 + 9)/16)`

= `7/25`

R.H.S cos²A - sin² A = `(4/5)^2 - (3/5)^2`

= `(16 - 9)/25`

= `7/25`