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Question
If sin A = `3/5` then show that 4 tan A + 3 sin A = 6 cos A
Solution
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)]
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