Question

Given: sec A = `( 29 )/(21), "evaluate : sin A" - 1/tan "A"`

Answer 1

Consider the diagram below :

sec A = `(29)/(21)`

i.e. `"hypotenuse"/"base" = (29)/(21) ⇒ "AC"/"AB" = (29)/(21)`

Therefore if length of AB = 21x , length of AC = 29x
Since

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

(21x)² + BC² = ( 29x)²

BC² = 841x² – 441x² =400x²

BC = 20x                    ...(perpendicular ) 

Now

sin A = `"perpendicular"/"hypotenuse" = (20x)/(29x) = 20/29`

tan A = `"perpendicular"/"base" = (20x)/(21x) = 20/21`

Therefore

sin A – `1/ tan "A"`

= `(20)/(29) – 1/(20/21)`

= `(20)/(29) –  21/20`

= – `(209)/(580)`