Question

Given: cos A = `( 5 )/ ( 13 )`

Evaluate:

1. `(sin "A "–cot "A") / (2 tan "A")`
2. `cot "A" + 1/cos"A"`

Answer 1

Consider the diagram below:

cos A = `( 5 )/( 13 )`

i.e. `"base"/"hypotenuse" = 5/13`

⇒ `"AB"/"AC" = 5/13`

Therefore, if length of AB = 5x, length of AC = 13x

Since

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

(5x)² + BC² = (13x)²

BC² = 169x² – 25x²

BC² = 144x²

∴ BC = 12x                 ...(perpendicular)

Now

tan A = `"perpendicular"/"base" = (12x)/(5x) = 12/5`

sin A =  `"perpendicular"/"hypotenuse"  = (12x)/(13x) = 12/13`

cot A = `"base"/"perpendicular" =(5x)/(12x) = 5/12`

(i) `(sin "A"  –  cot"A")/ (2tan "A")`

= `( 12 /13 – 5/12)/(2(12/5)`

= `(79)/(156). (5)/(24)`

= `( 395)/(3744 )`

(ii) cot A + `1/ cos "A"`

= `(5)/(12) + (1)/(5/13)`

= `(5)/(12) + (13)/(5)`

= `(181)/(60)`