Question

Given: 4 cot A = 3
find :
(i) sin A
(ii) sec A
(iii) cosec²A - cot²A.

Answer 1

Consider the diagram below :

4 cot A = 3

cot A = `(3)/(4)`

i.e.`"base"/"perpendicular" = (3)/(4) ⇒ "AB"/"BC" =(3)/(4)`

Therefore if length of AB = 3x, length of BC = 4x

Since
AB² + BC² = AC²             ...[ Using Pythagoras Theorem ]

(3x)² + (4x)² = AC²

AC² = 9x² + 16x² = 25x²

∴ AC = 5x  ...( hypotenuse )

(i) sin  A = `"perpendicular"/"hypotenuse " = (4x)/(5x) = (4)/(5)`

(ii) sec A = `"hypotenuse"/"base" = "AC"/"AB" = (5x)/(3x) = 5/3`

(iii) cosec A = `"hypotenuse"/"perpendicular" = "AC"/"BC" = (5x)/(4x) = (5)/(4)`

cot A = `(3)/(4)`

cosec² A – cot² A 

=`(5/4)^2 – (3/4)^2`

= `( 25 - 9)/(16)`

= `(16)/(16)`

= 1