Question

 If 4sinx=3cosx then find the value of (4sinx-cosx)/(4sinx+cosx)


answer it correctly please ..✌​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given:

$\tt\implies 4\: sin(x) = 3\: cos(x)$

$\tt\implies \dfrac{sin(x)}{cos(x)} =\dfrac{3}{4}\ \ \bigstar$

Therefore:

$\tt=\dfrac{4\: sin(x) - cos(x)}{4\:sin(x) + cos(x)}$

Dividing both numerator and denominator by cos(x), we get:

$\tt=\dfrac{4\: \dfrac{sin(x)}{cos(x)} - \dfrac{cos(x)}{cos(x)}}{4\: \dfrac{sin(x)}{cos(x)} +\dfrac{cos(x)}{cos(x)}}$

Substituting the value of sin(x)/cos(x), we get:

$\tt=\dfrac{4\cdot \dfrac{3}{4} -1}{4\cdot\dfrac{3}{4}+1}$

$\tt=\dfrac{3 -1}{3+1}$

$\tt=\dfrac{2}{4}$

$\tt=\dfrac{1}{2}$

Therefore:

$\dag\ \boxed{\tt\dfrac{4\: sin(x) - cos(x)}{4\:sin(x) + cos(x)}=\dfrac{1}{2}}$

$\texttt{\textsf{\large{\underline{Learn More}:}}}$

1. Relationship between sides.

1. sin(x) = Height/Hypotenuse.
2. cos(x) = Base/Hypotenuse.
3. tan(x) = Height/Base.
4. cot(x) = Base/Height.
5. sec(x) = Hypotenuse/Base.
6. cosec(x) = Hypotenuse/Height.

2. Square formulae.

1. sin²x + cos²x = 1.
2. cosec²x - cot²x = 1.
3. sec²x - tan²x = 1

3. Reciprocal Relationship.

1. sin(x) = 1/cosec(x).
2. cos(x) = 1/sec(x).
3. tan(x) = 1/cot(x).

4. Cofunction identities.

1. sin(90° - x) = cos(x) and vice versa.
2. cosec(90° - x) = sec(x) and vice versa.
3. tan(90° - x) = cot(x) and vice versa.

Answer 2

Given :- If 4sinx=3cosx then find the value of (4sinx-cosx)/(4sinx+cosx) ?

Solution :-

→ 4sin x = 3 cos x

→ sin x / cos x = 3/4

→ tan x = 3/4 = P/B

so,

→ H = √P² + B² = √(3² + 4²) = √(9 + 16) = √25 .

then,

→ sin x = P/H = 3/5

→ cos x = B/H = 4/5

putting both values we get,

→ (4sinx - cosx)/(4sinx + cosx)

→ [4 * (3/5) - (4/5)] / [4 * (3/5) + (4/5)]

→ (12/5 - 4/5) / (12/5 + 4/5)

→ (8/5) / (16/5)

→ (8/5) * (5/16)

→ (8/16)

→ (1/2) (Ans.)

Learn more :-

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