Question

If
$f(x) = ln(sinx) \: and \: g(x) = ln( cosx) \: then \: prove \: that \: {e}^{2g(a)} - {e}^{2f(a)} = {e}^{g(2a)}$
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Answer 1

Answer:

Given,

$\: \: \: \sf \: f(x) = ln \: (sin \: x) \\ \sf \therefore \: f(a) = ln \: (sin \: a) \\ \bf \: and \: \\ \: \: \: \sf \: g(x) = ln \: (cos \: x) \\ \sf \therefore \:g(a) = ln \: (cos \: a) \\ \sf \therefore \:g(2a) = ln \: (cos \: 2a) \\ \\ \bf \: now \\ L.H.S \sf \: = {e}^{2g(a)} - {e}^{2f(a)} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = {e}^{2 \: ln \: (cos \: a)} - {e}^{2 \: ln \: (sin \: a)} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = {e}^{ln \: ( {cos}^{2} \: a)} - {e}^{ln \:( {sin}^{2} \: a)} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = {cos}^{2} \: a - {sin}^{2} \: a \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = cos \: 2a \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = {e}^{ln \: (cos \: 2a)} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = {e}^{g(2a)} \\ \: \: \: \: \: \: \: \: \: \: \: \: = R.H.S \\ \\ \green{ \large {( \sf \: hence \: proved)}}$

Note :

$\bf \: = > a \: ln \: x = ln \: {x}^{a} \\ \\ \bf \: = > \: ln \: e = 1 \\ \\ \bf \: = > {e}^{ln \: x} \: = x \: ln \: e = x \times 1 = x \\ \\ \sf \: = > {cos}^{2} \theta - {sin}^{2} \theta = cos \: 2 \theta$

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Hi , ヽ(•‿•)ノ

Good morning.

Hope, you are fine by the grace of Allah.

And you are studying very well. d(✪‿✪)

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Alhamdulillah, I am also fine.