Question

(sinx-2sin^3x)=(2cos^3x-cosx)tanx
​

Answer 1

⭐Given⭐

$(\sin x - 2 \sin^{3} x) = \tan x(2 { \cos }^{3} x - \cos x)$

⭐To Find:⭐

Prove that LHS is equal to RHS.

⭐Answer:⭐

We will solve LHS and RHS one by one.

Firstly , we will solve LHS.

LHS:

$( \sin x - {\sin}^{3} x)$

Taking sin x as common, we get:

$\sin x(1 - 2 {\sin}^{2} x)$

We know that,

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

Substituting the above value in place of 1 , we get:

$\sin x( {\sin}^{2} x + {\cos}^{2} x - 2\sin ^{2} x)$

On Simplifying, we get:

$\sin x( {\cos}^{2} x - {\sin}^{2} x)$

Now, we will solve RHS.

RHS:

$\tan x(2 {\cos}^{3} x - \cos x)$

Taking cos x as common, we get:

$\tan x \times \cos x(2 {\cos}^{2} x - 1)$

We know that,

$\tan x = \frac{\sin x}{\cos x }$

Substituting the value of tan x in the equation, we get:

$\frac{\sin x}{\cos x} \times \cos x(2 {\cos}^{2} x - 1)$

We know that,

${\sin}^{2} x + {\cos}^{2} x = 1$

Substituting the value of 1 , we get:

$\sin x( 2{\cos}^{2} x - {\sin}^{2} x - {\cos}^{2} x)$

$\sin x( {\cos}^{2} x - {\sin}^{2} x)$

LHS = RHS

Hence Proved

⭐Other Trigonometric

Identities:⭐

$1) \: {\cosec}^{2} \theta - {\cot}^{2} \theta = 1$

$2) \: {\sec}^{2} \theta - {\tan}^{2} \theta = 1$

⭐Trigonometric Ratios:⭐

$1) \: \sin\theta = \frac{1}{ \cosec \theta} \\ 2) \: \cos \: \theta \: \: = \: \frac{1}{\sec\theta} \\ 3) \: \tan \: \theta \: = \: \frac{1}{\cot \: \theta} \\ 4) \: \cot \: \theta \: = \: \frac{1}{\tan \: \theta} \\ 5) \: \sec \: \theta \: = \: \frac{1}{\cos \: \theta} \\ 6) \cosec \theta= \frac{1}{\sin \: \theta}$