Question

If 3 cos A = 4 sin A, find the value of :
(i) cos A(ii) 3 - cot² A + cosec²A.

Answer 1

Consider the diagram below :

3cos A = 4 sin A

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

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

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

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

(4x)² + (3x)² = AC²

AC² = 25x²

∴ AC = 5x           ...( hypotenuse)

(i) cos A = `"AB"/"AC" = (4)/(5)`

(ii) cosec A = `"AC"/"BC" = (5)/(3)`

Therefore
3–cot² A + cosec² A

= `3  – (4/3)^2+(5/3)^2`

= `(27 –  16 + 25)/(9)`

=`(36)/(9)`

= 4