Question

1+ sin theta whole square + 1 minus sin theta whole square by 2 cos square theta is equal to 1 plus sin square theta by 1 minus sin square theta ​

Answer 1

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

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

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

$\mathsf{Consider,}$

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

$\textsf{Using the identities,}$

$\boxed{\begin{minipage}{5cm} \\\mathsf{(a+b)^2=a^2+b^2+2ab}\\\\\mathsf{(a-b)^2=a^2+b^2-2ab}\\ \end{minipage}}$

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

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

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

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

$\mathsf{=\dfrac{1+sin^2\theta}{cos^2\theta}}$

$\mathsf{Using,}\;\boxed{\mathsf{cos^2A=1-sin^2A}}$

$\mathsf{=\dfrac{1+sin^2\theta}{1-sin^2\theta}}$

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

Answer 2

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