Question

#QualityQuestion
@Trigonometry

If $\sf{2-cos^{2}\theta = 3sin\theta cos\theta}$, where $\sf{sin\theta ≠ cos\theta}$, the value of tanθ is?​

Answer 1

Answer: (theta is written as A)

1/2

Step-by-step explanation:

=> 2 - cos²A = 3sinAcosA

Divide both sides by cos²A :

=> (2 - cos²A)/cos²A = 3sinAcosA

=> 2/cos²A - cos²A/cos²A = 3sinAcosA/cosA

=> 2sec²A - 1 = 3tanA

=> 2(1 + tan²A) - 1 = 3 tanA

=> 2tan²A - 3tanA + 1 = 0

=> 2tan²A - 2tanA - tanA + 1 = 0

=> 2tanA(tanA - 1) - (tanA - 1) = 0

=> (tanA - 1)(2tanA - 1) = 0

=> tanA = 1 or tanA = 1/2

But for tanA = 1, sinA = cosA

So, tanA = 1/2

Line4: 1/cosA =secA,thus 2/cos²A=2sec²A

Line5: sec²A = 1 + tan²A

Line7: splitting the middle & then factorizing.

Answer 2

~Solution :-

The answer of your question will be $\sf \frac{1}{2}$

As,

$\sf \frac{(2 - co {s}^{2} A)}{co {s}^{2} A}= 3sinAcosA$

$\to \sf \: \frac{2}{co {s}^{2} A} - \frac{co {s}^{2} A}{co {s}^{2} A} = \frac{ 3sinAcosA}{cosA }$

$\to \sf 2se {c}^{2} A - 1 = 3tanA$

$\to \sf {2(1 + ta {n}^{2} A) - 1 = 3 tanA}$

$\to \sf{ 2ta {n}^{2} A - 3tanA + 1 = 0 }$

$\to \sf{2ta {n}^{2} A - 2tanA - tanA + 1 = 0}$

$\to \sf{ 2tanA(tanA - 1) - (tanA - 1) = 0}$

$\to \sf{ (tanA - 1)(2tanA - 1) = 0 }$

$\implies \bf \red{ tanA = \frac{ 1}{2}}$

• Thus, the value of $\tt{tanA =\frac{1}{2}}$.

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