Question

The value of cos2θ/1−sin2θ=​

Answer 1

$\frac{cos \: 2 \alpha }{1 - sin \: 2 \alpha } \\ \\ = \frac{ {cos} ^{2} \alpha \: - {sin}^{2} \alpha }{1 - 2sin \: \alpha \: cos \: \alpha } \\ \\ = \frac{(cos \: \alpha - sin \: \alpha )(cos \: \alpha + \: sin \: \alpha )}{ {cos}^{2} \alpha + {sin}^{2} \alpha - 2sin \: \alpha \: cos \: \alpha } \\ \\ = \frac{(cos \: \alpha - sin \: \alpha )(cos \: \alpha + \: sin \: \alpha )}{ {(cos \: \alpha - \: sin \: \alpha )}^{2} } \\ \\ = \frac{cos \: \alpha + sin \: \alpha }{cos \: \alpha - \: sin \: \alpha } \\ \\ = \frac{1 + \: tan \: \alpha }{1 - tan \: \alpha } \\ \\ = tan \: (45 + \alpha )$

Answer 2

ANSWER:

To Find:

• Value of (cos2θ)/(1−sin2θ)

Solution:

$:\longrightarrow \dfrac{\cos2\theta}{1-\sin2\theta}\\\\\text{We know that, \cos2\theta=\cos^2\theta-\sin^2\theta and \sin2\theta=2\sin\theta\cos\theta }\\\\:\implies\dfrac{\cos^2\theta-\sin^2\theta}{1-2\sin\theta\cos\theta}\\\\\text{Also, a^2-b^2=(a+b)(a-b) and 1=\sin^2\theta+\cos^2\theta }\\\\:\implies\dfrac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta} \\\\:\implies\dfrac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{\cos^2\theta+\sin^2\theta-2\cos\theta\sin\theta}$

$\text{We know that, a^2+b^2-2ab=(a-b)^2 . So,}\\\\:\implies\dfrac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)^2}\\\\:\implies\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\\\\\text{Dividing both numerator and denomimator by \cos\theta }\\\\\:\implies\dfrac{\dfrac{\cos\theta+\sin\theta}{\cos\theta}}{\dfrac{\cos\theta-\sin\theta}{\cos\theta}}\\\\:\implies\dfrac{\dfrac{\cos\theta}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}}{\dfrac{\cos\theta}{\cos\theta}-\dfrac{\sin\theta}{\cos\theta}}$

$\text{Also, \dfrac{\sin\theta}{\cos\theta}=\tan\theta }\\\\:\implies\dfrac{1+\tan\theta}{1-\tan\theta}\\\\:\implies\dfrac{1+\tan\theta}{1-\tan\theta\times(1)}\\\\\text{We know that, \tan\left(\dfrac{\pi}{4}\right)=1 . So,} \\\\:\implies\dfrac{\tan\left(\dfrac{\pi}{4}\right)+\tan\theta}{1-\left(\tan\theta\times\tan\left(\dfrac{\pi}{4}\right)\right)}\\\\\text{We know that, \dfrac{\tan A+\tan B}{1-(\tan A\times\tan B)} = \tan(A+B) . So,}\\\\\bf{:\implies \tan\left(\dfrac{\pi}{4}+\theta\right)}$

Formulae Used:

• cos2θ = cos²θ - sin²θ
• sin2θ = 2sinθcosθ
• a² - b² = (a + b)(a - b)
• 1 = sin²θ + cos²θ
• a² + b² - 2ab = (a - b)²
• (tanA+tanB)/(1-tanA*tanB) = tan(A + B)