Question

sin tetha ― 2sin^3 tetha/2cos^3tetha ― costetha = tan tetha how​

Answer 1

Question :

To prove that

$\sf \dfrac{ \sin \theta - 2 \sin {}^{3} \theta }{2 \cos {}^{3} \theta - 1} = \tan \theta$

Formula's Used:

$\sf1) \cos2 \theta = 2 \cos {}^{2} \theta - 1$

$2) \cos2 \theta = 1 - 2 \sin {}^{2} \theta$

$\sf3) \sin2 \theta = 2 \sin \theta \cos \theta$

$\sf4) \cos2 \theta = \cos {}^{2} \theta - \sin {}^{2} \theta$

Solution :

We have to prove that

$\sf \dfrac{ \sin \theta - 2 \sin {}^{3} \theta }{2 \cos {}^{3} \theta - \cos \theta } = \tan \theta$

LHS $\sf=\dfrac{ \sin \theta - 2 \sin {}^{3} \theta }{2 \cos {}^{3} \theta - \cos \theta }$

$\sf = \dfrac{ \sin \theta(1 - 2 \sin {}^{2} \theta)}{ \cos \theta(2 \cos {}^{2} \theta - 1)}$

$\sf = \tan \theta \times \dfrac{1 - 2 \sin {}^{2} \theta}{2 \cos {}^{2} \theta - 1 }$

Use Formula cos2θ=2cos²θ-1 or 1-2sin²θ

$\sf= \tan \theta \times \dfrac{ \cos2 \theta}{ \cos2 \theta}$

$\sf = \tan \theta$

RHS =tanθ

⇒LHS=RHS

Hence proved

Answer 2

To Prove :

$\tt\dagger \: \: \: \: \: \dfrac{ sin \theta - 2 {sin}^{ 3} \theta}{2{cos }^{3} \theta - 1 } = tan \theta.$

Solution :

Taking LHS,

$\tt\longmapsto \dfrac{ sin \theta - 2 {sin}^{ 3} \theta}{2{cos }^{3} \theta - 1 }$

$\tt\longmapsto \dfrac{ sin \theta (1- 2 {sin}^{ 2} \theta)}{cos \theta(2{cos }^{2} \theta - 1) }$

$\tt\longmapsto \dfrac{ sin \theta [1- 2(1 - {cos}^{ 2} \theta)]}{cos \theta(2{cos }^{2} \theta - 1) }$

$\tt\longmapsto \dfrac{ sin \theta (1- 2 + 2 {cos}^{ 2} \theta)}{cos \theta(2{cos }^{2} \theta - 1) }$

$\tt\longmapsto \dfrac{ sin \theta (2 {cos}^{2} \theta + 1 - 2) }{cos \theta(2{cos }^{2} \theta - 1) }$

$\tt\longmapsto \dfrac{ sin \theta \cancel{(2 {cos}^{2} \theta - 1 ) }}{cos \theta\cancel{(2{cos }^{2} \theta - 1) }}$

$\tt\longmapsto \dfrac{ sin \theta }{cos \theta}$

$\tt\longmapsto tan \theta.$

Hance LHS = RHS.

$\rule{200}3$

$\boxed{\begin{minipage}{6 cm}Fundamental Trigonometric Identities \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\1+\tan^2\theta = \sec^2\theta \\ \\ \sin^2\theta = 1 - \cos^2\theta \\ \\1+\cot^2\theta = \text{cosec}^2 \, \theta \end{minipage}}$