Question

please solve this
10 maths
board exam paper

Answer 1

Hey there !

Solution:

$\dfrac{ SinA - 2\:Sin^3A}{2\:Cos^3A - Cos A } = Tan A \\ \\ \text{ Taking common terms out, we get,} \\ \\ \implies \dfrac{ Sin A ( 1 - 2\:Sin^2A) }{ CosA ( 2\:Cos^2A - 1)} \\ \\ \\ \text{ We know that,:} \\ \\ Cos \:2A = 1 - 2\:Sin^2A \\ \\ Cos 2A = 2\:Cos^2A - 1 \\ \\ \text{Hence we get,} \\ \\ \\ \implies \dfrac{ SinA ( Cos 2A )}{Cos A ( Cos2A )} \\ \\ \\ Cos\:2A \:\: gets \:\: cancelled. Hence, \\ \\ \\ \implies \dfrac{ Sin A}{CosA } = Tan A \\ \\ \\ \textbf{Hence Proved }$

Hope my answer helped !

Answer 2

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R.H.S.

$\frac{ sin \: A +2 \: sin^{3}\:A}{2\:cos^{3}A-cos \: A}\\\\\frac{sin\: A(1-2\: sin^{2}\:A)}{cos\:A(2cos^{2}\:A-1)}\\\\ \frac{tan\: A(1-2\: sin^{2}\:A) }{(2cos^{2}\:A-1)}$

$\therefore\:cos^{2}\: A + sin^{2}\: A= 1\\ cos^{2}\:A-1 =\: -sin^{2}\: A\\ or \\ cos^{2} \: A = 1-sin^{2}\:A$

........(equation 1)

Now,

$tan\: A \: \frac{[(1\: - sin^{2} \:A )- sin^{2}\:A)}{[cos^{2}\:A +(cos^{2}\: A - 1)]}\\ \\tan\: A\: \frac{(cos^{2}\:A - sin^{2} \:A)}{(cos^{2}\: A - sin^{2}\:A)}$

$tan\: A \: \frac{ \cancel{(cos^{2}\: A - sin^{2} \: A)}}{\cancel{(cos^{2} \: A - sin^{2} \:A)}}\\\\ tan \: A$

R.H.S

L.H.S = R.H.S

Hence Proved