Question

sinA - 2sin³A÷2cos³A-cosA=tanA
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Answer 1

Answer:

It is given that,

sinA-2sin³A/2cos³A-cosA

=SinA (1-2sin²A)/cosA (2cos²A-1)

As we know,

Cos2A=1-2sin²A=2cos²A-1

=SinA (Cos2A)/cosA (Cos2A)

=SinA/cosA

=tan A

Step-by-step explanation:

hope it helps you frnd..

Answer 2

GIVEN:

$\dfrac{sin \: A - 2sin^{3} \: A}{2cos^{3} A-cos \: A}=tan \: A$

TO PROVE:

$\dfrac{sin \: A - 2sin^{3} \: A}{2cos^{3} A-cos \: A}=tan \: A$

EXPLANATION:

$\dfrac{sin \: A - 2sin^{3} \: A}{2cos^{3} A-cos \: A}=tan \: A$

$\dfrac{sin\: A(1 - 2sin^{2} \: A)}{cos\: A(2cos^{2} A-1)}=tan \: A$

$\boxed{ \large{ \bold{ \frac{sin\: A}{cos\: A} = tan\: A }}}$

$\boxed{ \large{ \bold{ 1 - {sin}^{2} \: A = {cos}^{2}\: A }}}$

$tan\: A \left(\dfrac{1 - 2sin^{2} \: A}{2(1 -{sin}^{2} \: A )-1}\right)=tan \: A$

$tan\: A \left(\dfrac{1 - 2sin^{2} \: A}{2-2{sin}^{2} \: A -1} \right)=tan \: A$

$tan\: A \left(\dfrac{1 - 2sin^{2} \: A}{1-2{sin}^{2} \: A} \right)=tan \: A$

$tan\: A =tan \: A$

L.H.S = R.H.S

HENCE PROVED.