Question

If sin θ = 3/5 find the value of 4 tanθ + 3 sinθ – 6 cosθ

Answer 1

`\boxed{\sf 0}`

Given:

`\sf sin \theta = (3)/(5)`

To Find:

`\sf 4tan \theta + 3sin \theta - 6cos \theta`

Explanation:

`\sf As \ we \ know, \\ \sf sin 37 \degree = (3)/(5) \\ \\ \sf \therefore \ sin \theta = sin 37 \degree \\ \\ \sf \implies \theta = 37 \degree`

`\sf \implies tan \theta = (3)/(4) \\ \\ \sf \implies cos \theta = (4)/(5)</p><p>`

`\sf So, \\ \sf \implies 4tan \theta + 3sin \theta - 6cos \theta \\ \\ \sf Putting \ the \ values \ of \ tan \theta , sin \theta \ and \ cos \theta \ respectively: \\ \sf \implies ( 4 * (3)/( 4)) + (3 * (3)/(5)) - (6 * (4)/(5) ) \\ \\ 4 * (3)/( 4) = 3: \\ \sf \implies \boxed{3} + (3 *(3)/(5)) - (6 * (4)/(5) ) \\ \\ \sf 3 * 3 = 9: \\ 3 + \frac{\boxed{9}}{5} - (6 * (4)/(5) ) \\ \\ \sf 6 * 4 = 24: \\ \sf 3 + (9)/(5) - \frac{\boxed{24}}{5} \\ \\ \sf Put \ 3 + (9)/(5) - (24)/(5) \ over \ the \ common \ denominator \ 5: \\ \sf \implies 3 * (5)/(5) + (9)/(5) - (24)/(5) \\ \\ \sf \implies (15)/(5) + (9)/(5) - (24)/(5) \\ \\\sf \implies (15 + 9 - 24)/(5) \\ \\ \sf 15 + 9 = 24: \\ \sf \implies \frac{\boxed{24} - 24}{5} \\ \\ \sf 24 - 24 = 0: \\ \sf \implies \frac{\boxed{0}}{5} \\ \\ \sf \implies 0`

Answer 2

Explanation:

If sine theta is 3/5, taking the sine inverse will give you 36.87°.

4tan theta is 4 × tan36.87 = 3

3sin theta = 3 × sin36.87 = 1.8

6cos theta is 6× cos36.87 = 4.8

Therefore, it becomes 3 + 1.8 - 4.8 = 4.8 - 4.8

The answer is 0