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 = `(3"k")/(4"k") = 3/4`

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

∴ cos A = `(4"k")/(5"k") = 4/5`

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

= `3 + 9/5`

=`(15 + 9)/5`

= `24/5`   ......(iii)

6cos A = `6(4/5) = 24/5`   ......(iv)

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