Question

hey #Quality Question
@Trigonometry

If θ is Acute angle and 3sinθ = 4cosθ then Find out the value of 4sin2θ - 3cos2θ + 2.
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Answer 1

Given :-

If θ is angle and 3sinθ = 4cosθ

To Find :-

the value of 4sin²θ - 3cos²θ + 2.

​Solution :-

3 sinθ = 4 cosθ

4/3 = sinθ/cosθ

Now, We know that

sinθ/cosθ = tanθ

4/3 = tanθ

On squaring both sides we get

(4/3)² = (tanθ)²

16/9 = tan²θ

sec²θ = 1 + tan²θ

sec²θ = 1 + 16/9

sec²θ = 9 + 16/9

sec²θ = 25/9

On rooting both sides

√(sin²θ) = √(25/9)

sinθ = 5/3

Now

cosθ = 1/sinθ

cosθ = 1/(5/3)

cosθ = 3/5

Now

sinθ = √(1 - cos²θ)

sinθ = √[1 - (3/5)²]

sinθ = √[1 - 9/25]

sinθ = √[25 - 9/25]

sinθ = √[16/25]

sinθ = 4/5

Finding the value

4 × (4/5)² - 3 × (3/5)² + 2

4 × (16/25) - 3 × (9/25) + 2

64/25 - 27/25 + 2

87/25

Answer 2

Answer:

Given:-

3sinθ = 4cosθ

$\frac{sinθ}{cosθ} = \frac{4}{3}$

$tanθ = \frac{4}{3}$

By Pythagoras theorem

$ac {}^{2} = ab {}^{2} + bc {}^{2}$

$ac {}^{2} = (4) {}^{2} + (3) {}^{2}$

$ac {}^{2} = 16 + 9$

$ac {}^{2} = 25$

$ac = \sqrt{25} = 5$

$ac = 5$

$\sinθ = \frac{ab}{ac} = \frac{4}{5}$

$\cosθ = \frac{bc}{ac} = \frac{3}{5}$

Given that,

4sin^2 θ - 3cos^2 θ + 2

$4 \times ( \frac{4}{5} ) {}^{2} - 3 \times ( \frac{3}{5}) {}^{2} + 2$

$\frac{64}{25} - \frac{27}{25} + \frac{50}{25}$

$= \frac{87}{25} answer$

hope it helps you