Prove the trigonometric identity: $\frac{\sin heta - 2 \sin^3 heta}{2 \cos^3 heta - \cos heta} = an heta$. Show all steps clearly.

Q: Prove the trigonometric identity: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$. Show all steps clearly.

To prove: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$ \nTake the Left Hand Side (LHS): $\text{LHS} = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$ \nFactor out $\sin \theta$ from the numerator and $\cos \theta$ from the denominator: $\text{LHS} = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$ \nUsing the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we can substitute $1 = \sin^2 \theta + \cos^2 \theta$ in the brackets, or use $\cos^2 \theta = 1 - \sin^2 \theta$:\nLet us substitute $\cos^2 \theta = 1 - \sin^2 \theta$ in the denominator:\nDenominator term $= 2(1 - \sin^2 \theta) - 1$ $= 2 - 2 \sin^2 \theta - 1$ $= 1 - 2 \sin^2 \theta$ \nNow, substitute this back into the LHS expression: $\text{LHS} = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (1 - 2 \sin^2 \theta)}$ \nCancel out the common term $(1 - 2 \sin^2 \theta)$ from both numerator and denominator: $\text{LHS} = \frac{\sin \theta}{\cos \theta}$ \nWe know that $\frac{\sin \theta}{\cos \theta} = \tan \theta$, which is equal to the Right Hand Side (RHS). $\text{LHS} = \text{RHS}$\nHence proved.

Question

Prove the trigonometric identity: $\frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta} = 	an 	heta$. Show all steps clearly.

Accepted Answer

To prove: $\frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta} = 	an 	heta$ 
Take the Left Hand Side (LHS): $	ext{LHS} = \frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta}$ 
Factor out $\sin 	heta$ from the numerator and $\cos 	heta$ from the denominator: $	ext{LHS} = \frac{\sin 	heta (1 - 2 \sin^2 	heta)}{\cos 	heta (2 \cos^2 	heta - 1)}$ 
Using the fundamental trigonometric identity $\sin^2 	heta + \cos^2 	heta = 1$, we can substitute $1 = \sin^2 	heta + \cos^2 	heta$ in the brackets, or use $\cos^2 	heta = 1 - \sin^2 	heta$:
Let us substitute $\cos^2 	heta = 1 - \sin^2 	heta$ in the denominator:
Denominator term $= 2(1 - \sin^2 	heta) - 1$ $= 2 - 2 \sin^2 	heta - 1$ $= 1 - 2 \sin^2 	heta$ 
Now, substitute this back into the LHS expression: $	ext{LHS} = \frac{\sin 	heta (1 - 2 \sin^2 	heta)}{\cos 	heta (1 - 2 \sin^2 	heta)}$ 
Cancel out the common term $(1 - 2 \sin^2 	heta)$ from both numerator and denominator: $	ext{LHS} = \frac{\sin 	heta}{\cos 	heta}$ 
We know that $\frac{\sin 	heta}{\cos 	heta} = 	an 	heta$, which is equal to the Right Hand Side (RHS). $	ext{LHS} = 	ext{RHS}$
Hence proved.

MP Board · Class 10 · Mathematics · Introduction to TrigonometryProve the trigonometric identity: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$. Show all steps clearly.