Question

Prove that : (Sin θ - 2Sin³ θ) / (2 Cos³ θ-Cos θ) = tan θ
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Answer 1

Step-by-step explanation:

sin - sin³ - sin³ / cos³ + cos³ - cos

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

sin (cos² - sin² ) / cos (cos² - ( 1 - cos² ))

sin (cos² - sin² ) / cos (cos² - sin² )

sin/cos = tan

L.H.S = R.H.S

Hence proved

Answer 2

Step-by-step explanation:

$\mathsf{Given : \dfrac{Sin\theta - 2Sin^3\theta}{2Cos^3\theta - Cos\theta}}$

$\mathsf{Taking\;Sin\theta\;Common\;in\;the\;Numerator\;and\;Cos\theta\;Common\;in\;the\;Denominator :}$

$\mathsf{\implies \dfrac{Sin\theta(1 - 2Sin^2\theta)}{Cos\theta(2Cos^2\theta - 1)}}$

$\mathsf{\implies (\dfrac{Sin\theta}{Cos\theta}) \dfrac{(1 - 2Sin^2\theta)}{(2Cos^2\theta - 1)}}$

$\mathsf{We\;know\;that :}$

✿  $\mathsf{Tan\theta = \dfrac{Sin\theta}{Cos\theta}}$

✿  $\mathsf{Cos2\theta = 1 - 2Sin^2\theta}$

✿  $\mathsf{Cos2\theta = 2Cos^2\theta - 1}$

$\mathsf{Substituting\;all\;the\;Respective\;Values,\;We\;get :}$

$\mathsf{\implies (Tan\theta) \dfrac{(Cos2\theta)}{(Cos2\theta)}}$

$\mathsf{\implies Tan\theta}$