Question

2 cos squared theta minus 1 to the power whole square upon cos to the power 4 theta minus sin to the power 4 theta equals to 1 - 2 sin square theta

Answer 1

$\textbf{To prove:}$

$\dfrac{(2\,cos^2\theta-1)^2}{cos^4\theta-sin^4\theta}=1-2\,sin^2\theta$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{(2\,cos^2\theta-1)^2}{cos^4\theta-sin^4\theta}$

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

$\text{Using the identity,}$

$\boxed{\bf\,a^2-b^2=(a-b)(a+b)}$

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

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

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

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

$=2\,cos^2\theta-1$

$=2(1-sin^2\theta)-1$

$=2-2\,sin^2\theta-1$

$=1-2\,sin^2\theta$

$\implies\boxed{\bf\dfrac{(2\,cos^2\theta-1)^2}{cos^4\theta-sin^4\theta}=1-2\,sin^2\theta}$

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Answer 2

Step-by-step explanation:

HOPE THIS HELPS YOU.

JAY SHRI KRISHNA