Question

how to prove lhs=rhs​

Answer 1

plz find the attached image.

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Answer 2

Answer:

$\implies \bf (Sin^{2} \: A)(Cos^{2} \: B) - (Cos^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

Using Identities :

$\bullet \: \: {\boxed{\sf {1 - Sin^{2} \: \theta = Cos^{2} \: \theta }}}$

$\bullet \: \: {\boxed{\sf {Sin^{2} \: \theta = 1 - Cos^{2} \: \theta}}}$

Let's solve it :

$\implies \sf (Sin^{2} \: A)(Cos^{2} \: B) - (Cos^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

$\bullet \: \: {\sf {Sin^{2} \: \theta = 1 - Cos^{2} \: \theta}}$

$\implies \sf (Sin^{2} \: A)(1 - Sin^{2} \: B) - (Cos^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

$\implies \sf (Sin^{2} \: A)(1) - (Sin^{2} \: A)(Sin^{2} \: B) - (Cos^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

$\implies \sf (Sin^{2} \: A) - (Sin^{2} \: A)(Sin^{2} \: B) - (Cos^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

$\implies \sf (Sin^{2} \: A) - (Sin^{2} \: A)(Sin^{2} \: B) - (1 - Sin^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

$\implies \sf (Sin^{2} \: A) - (Sin^{2} \: A)(Sin^{2} \: B) - (1)(Sin^{2} \: B) - (Sin^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

$\implies \sf (Sin^{2} \: A) - (Sin^{2} \: A)(Sin^{2} \: B) - (Sin^{2} \: B) - (Sin^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

$\implies \sf (Sin^{2} \: A) - (Sin^{2} \: A)(Sin^{2} \: B) - (Sin^{2} \: B) + (Sin^{2} \: A)(Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

$\implies \sf (Sin^{2} \: A) - {\not{(Sin^{2} \: A)(Sin^{2} \: B)}} - (Sin^{2} \: B) + {\not(Sin^{2} \: A)(Sin^{2} \: B)} = Sin^{2} \: A - Sin^{2} \: B$

$\implies \sf (Sin^{2} \: A) - (Sin^{2} \: B) = Sin^{2} \: A - Sin^{2} \: B$

${\boxed{\bf{L.H.S = R.H.S}}}$

Verified !!