Question

Can Anyone Solve it..... ✌..... ╮(╯▽╰)╭​

Answer 1

Answer:

(sin@-2sin³@)/(2cos³@-cos@)

=sin@(1-2sin²@)/cos@(2cos²@-1)

=sin@(1-sin²@-sin²@)/cos@(cos²@+cos²@-1)

=tan@{(1-sin²@-sin²@)/(cos²@+cos²@-1)}

=tan@{(cos²@-sin²@)/(cos²@-sin²@)}

.....1-sin²@=cos²@&cos²@-1=-sin²@

=tan@(1)

=tan@

hence proved

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Answer 2

Answer:

Let angle is

$alpha( \alpha )$

To prove :

$\frac{sin \alpha - 2 {sin}^{3} \alpha }{2 {cos}^{3} \alpha -cos \alpha } = tan \alpha$

Proof :

LHS

$= \frac{sin \alpha - 2 {sin}^{3} \alpha }{2 {cos}^{3} \alpha -cos \alpha } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha )}{cos \alpha (2 {cos}^{2} \alpha - 1) } \\ \\ = tan \alpha \times \frac{(1 - {sin}^{2} \alpha - {sin}^{2} \alpha ) }{( {cos}^{2} \alpha + {cos}^{2} \alpha - 1)} \\ \\ = tan \alpha \times \frac{( {cos}^{2} \alpha - {sin}^{2} \alpha ) }{( {cos}^{2} \alpha - {sin}^{2} \alpha )} \\ \\ = tan \alpha \times 1 \\ \\ = tan \alpha$

= RHS

Therefore, LHS = RHS

Hence Proved.

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