Question

✌❤ ₱Lz SøLv€ ¡t .....✌❤​

Answer 1

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$Prove: \\ cos^2\theta . cos^2\beta - sin^2 \theta . sin^2 \beta = cos^\theta - sin^2 \beta$

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$\boxed{\red{\bold{Solving\:LHS:}}}$

$\purple{\implies cos^2\theta . cos^2 \beta - sin^2 \theta . sin^2 \beta}$

$\purple{\implies}cos^2 \theta (1-sin^2 \beta) - (1-cos^2 \theta)sin^2 \beta$

$\blue{(\because sin^2 \Phi + cos^2 \Phi = 1)}$

$\green{\implies cos^2 \theta - cos^2 \theta sin^2\beta -sin ^2\beta + cos^2 \theta sin^2 \beta}$

$\green{\implies{\boxed{\bold{cos^2 \theta - sin^2 \beta = RHS}}}}$

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Answer 2

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$\boxed{ \red{ \mathfrak{Solving \: L.H.S. : }}}$

$\\ \\ \bf{Formulas \: used:-} \\ \boxed{ \bf{ {sin}^{2}A + {cos}^{2}A = 1 }} \\ \\ ⟹ \orange{\bf{ {cos}^{2} \theta. {cos}^{2} \beta - {sin}^{2} \theta. {sin}^{2} \beta } } \\ ⟹ \bf{ {cos}^{2} \theta.(1 - {sin}^{2} \beta ) - (1 - {cos}^{2} \theta). {sin}^{2} \beta } \\ ⟹ \bf{ {cos}^{2} \theta - {cos}^{2} \theta. {sin}^{2} \beta - {sin}^{2} \beta + {cos}^{2} \theta. {sin}^{2} \beta } \\ ⟹ { \orange{ \boxed{ \bf{ \purple{ {cos}^{2} \theta - {sin}^{2} \beta = R.H.S. } }}}}$