Question

If sin A = `3/5`, then show that 4 tan A + 3 sin A = 6 cos A.

Answer 1

sin A = `3/5`   ...(i) [Given]

In ∆ABC,

Let ∠ABC = 90°

∴ `sin A = (BC)/(AC)`   ...(ii) [By definition]

∴ `(BC)/(AC) = 3/5`   ...[From (i) and (ii)]

Let BC = 3k, AC = 5k

In ∆ABC, ∠B = 90°

∴ AB² + BC² = AC²    ...[Pythagoras theorem]

∴ AB² + (3k)² = (5k)²

∴ AB² + 9k² = 25k²

∴ AB² = 25k² – 9k²

∴ AB² = 16k² 

∴ AB = 4k   ...[Taking square root of both sides]

Now, `tan A = (BC)/(AB)`   ...[By definition]

∴ `tan A = (3k)/(4k) = 3/4`

`cos A = (AB)/(AC)`   ...[By definition]

∴ `cos A = (4k)/(5k) = 4/5`

∴ `4 tan A + 3 sin A = 4(3/4) + 3(3/5)`

= `3 + 9/5`

=`(15 + 9)/5`

= `24/5`   ...(iii)

`6 cos A = 6(4/5)` 

= `24/5`   ...(iv)

∴ 4 tan A + 3 sin A = 6 cos A   ...[From (iii) and (iv)]