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seraphim [82]
3 years ago
5

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

Mathematics
2 answers:
julsineya [31]3 years ago
7 0

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

zepelin [54]3 years ago
5 0

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

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