Question

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

Answer 1

Answer:

$\boxed{\sf 0}$

Given:

$\sf sin \theta = \frac{3}{5}$

To Find:

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

Step-by-step explanation:

$\sf As \ we \ know, \\ \sf sin 37 \degree = \frac{3}{5} \\ \\ \sf \therefore \ sin \theta = sin 37 \degree \\ \\ \sf \implies \theta = 37 \degree$

$\sf \implies tan \theta = \frac{3}{4} \\ \\ \sf \implies cos \theta = \frac{4}{5}$

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

Answer 2

Step-by-step 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