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" angle A)/("Side opposite to"angle 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"angleA)/"Hypotenuse" = (AB)/(AC) `

= 4k/5k = 4/5

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

=3k/5k = 3/5

`tan A = ("Side adjacent to"angleA)/"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`

cos² A + sin² 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 = 25

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 cos^2A - sin^2 A = (4/5)^2 - (3/5)^2 = (16 - 9)/25`

`= 7/25`