We know that
\( tan2A=\dfrac{2tanA}{1-tan^2A} \)
L.H.S. = tan4x
=tan2(2x)
\(\dfrac{2tan2x}{1-tan^2(2x)} \)
\( \dfrac{\dfrac{4tanx}{1-tan^2x}}{[1-\dfrac{4tan^2x}{(1-tan^2x})^2]}\)
\( tan4x=\dfrac{4tanx(1-tan^2x)}{1-6tan^2x+tan^4x}\)
Further computing, we obtain
= \( \dfrac{\dfrac{4tanx}{1-tan^2x}}{[\dfrac{(1-tan^2x)^24tan^2x)}{(1-tan^2x)^2}]}\)
= \( \dfrac{4tanx(1-tan^2x)}{1+tan^4x-2tan^2x-4tan^2x}\)
= \( \dfrac{4tanx(1-tan^2x)}{1-6tan^2x+tan^4x}\)
= R.H.S.
Hence proved.
Answered by Pragya Singh | 1 year agoprove that \(sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}\)
prove that \( 3 sin \dfrac{π}{6} sec \dfrac{π}{3} – 4 sin \dfrac{5π}{6} cot \dfrac{π}{4} = 1\)
prove that \( tan \dfrac{11π}{3} – 2 sin \dfrac{4π}{6} – \dfrac{3}{4} cosec^2 \dfrac{π}{4} + 4 cos^2 \dfrac{17π}{6} = \dfrac{(3 – 4\sqrt{3})}{2}\)
prove that cos 570° sin 510° + sin (-330°) cos (-390°) = 0
prove that tan (-125°) cot (-405°) – tan (-765°) cot (675°) = 0