Question

1 - cos2 theta + sin2 theta ÷ 1 + cos 2 theta sin2 theta = tan theta

Answer 1

$\underline{\textbf{To prove:}}$

$\mathsf{\dfrac{1-cos\,2\theta+sin\,2\theta}{1+cos\,2\theta+sin\,2\theta}=tan\,\theta}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{1-cos\,2\theta+sin\,2\theta}{1+cos\,2\theta+sin\,2\theta}}$

$\mathsf{=\dfrac{1-(1-2\,sin^2\theta)+2\,sin\,\theta\,cos\,\theta}{1+(2\,cos^2\theta-1)+2\,sin\,\theta\;cos\,\theta}}$

$\mathsf{=\dfrac{2\,sin^2\theta+2\,sin\,\theta\,cos\,\theta}{2\,cos^2\theta+2\,sin\,\theta\;cos\,\theta}}$

$\mathsf{=\dfrac{2\,sin\theta(sin\,\theta+cos\,\theta)}{2\,cos\theta(cos\,\theta+sin\,\theta)}}$

$\mathsf{=\dfrac{sin\theta(sin\,\theta+cos\,\theta)}{cos\theta(sin\,\theta+cos\,\theta)}}$

$\mathsf{=\dfrac{sin\theta}{cos\theta}}$

$\mathsf{=tan\,\theta}$

$\implies\boxed{\bf\dfrac{1-cos\,2\theta+sin\,2\theta}{1+cos\,2\theta+sin\,2\theta}=tan\,\theta}$

$\underline{\textbf{Formula used:}}$

$\boxed{\begin{minipage}{4cm} \\\mathsf{cos\,2A=1-2\,sin^2A}\\\\\mathsf{cos\,2A=2\,cos^2A-1}\\ \end{minipage}}$