Question

If (sin θ + cos θ)/( sinθ - cos θ) = 3/2, then the value of
tan θ =
with explanation

Answer 1

Answer:

tanθ = 5.

Step-by-step explanation:

ATQ,

$\Longrightarrow \sf \dfrac{sin\theta+cos\theta}{sin\theta-cos\theta} = \dfrac{3}{2}$

On cross-multiplying we get,

$\Longrightarrow \sf 2(sin\theta+cos\theta) = 3(sin\theta-cos\theta)$

$\Longrightarrow \sf 2sin\theta+2cos\theta=3sin\theta-3cos\theta$

$\Longrightarrow \sf 2sin\theta-3sin\theta=-3cos\theta-2cos\theta$

$\Longrightarrow \sf -sin\theta=-5cos\theta$

Cancelling the negative sign we get,

$\Longrightarrow \sf \ sin\theta=5cos\theta$

Transposing cosθ to other side of the equation we get,

$\Longrightarrow \sf \dfrac{sin\theta}{cos\theta}=5$

$\sf We \ know \ that \ \dfrac{sin\theta}{cos\theta}=tan\theta$

$\Longrightarrow \sf tan\theta = 5$

∴ The value of tanθ = 5.

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Cross-checking the answer:

$\sf ATQ, \ \sf \dfrac{sin\theta+cos\theta}{sin\theta-cos\theta} = \dfrac{3}{2}$

$\sf LHS = \dfrac{sin\theta+cos\theta}{sin\theta-cos\theta}$

Dividing the numerator and denominator by cosθ we get,

$\sf LHS = \dfrac{\frac{sin\theta}{cos\theta}+\frac{cos\theta}{cos\theta}}{\frac{sin\theta}{cos\theta}-\frac{cos\theta}{cos\theta}}$

$\sf LHS = \dfrac{tan\theta+1}{tan\theta-1}$

Substituting the value of tanθ above we get,

$\sf LHS = \dfrac{5+1}{5-1}$

$\sf LHS = \dfrac{6}{4}$

$\sf LHS = \dfrac{3}{2}$

$\sf LHS = RHS$

Hence the answer is correct.

Answer 2

Step-by-step explanation:

Given:

$\huge{\sf {\frac{sinθ+cosθ}{sinθ-cosθ}}}$ = $\huge{\sf {\frac{3}{2}}}$

To Find:

• Value of tanθ

Solution: Doing cross multiplication

→ 2(sinθ + cosθ) = 3( sinθ – cosθ)

→ 2sinθ + 2cosθ = 3sinθ – 3cosθ

→ 2cosθ + 3cosθ = 3sinθ – 2sinθ ..( Transport similar terms from RHS to LHS or LHS to RHS )

→ 5cosθ = sinθ

→ 5 = $\</strong><strong>h</strong><strong>u</strong><strong>g</strong><strong>e{\sf {\frac{</strong><strong>sinθ</strong><strong>}{</strong><strong>cosθ</strong><strong>}}}$

Since, $\huge{\sf {\frac{sinθ}{cosθ}}}$ = tanθ

→ 5 = tanθ

Hence, The value of tanθ will be 5