LHS $= \dfrac{\sin A - 2\sin^3 A}{2\cos^3 A - \cos A} + \dfrac{\cos A}{\sin A}$
Step 1: Factorise the first term:
$$= \frac{\sin A(1 - 2\sin^2 A)}{\cos A(2\cos^2 A - 1)} + \frac{\cos A}{\sin A}$$
Since $1 - 2\sin^2 A = 2\cos^2 A - 1$ (using $\sin^2 A + \cos^2 A = 1$):
$$= \frac{\sin A(2\cos^2 A - 1)}{\cos A(2\cos^2 A - 1)} + \frac{\cos A}{\sin A}$$
$$= \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}$$
Step 2: Combine:
$$= \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$$
$$= \frac{1}{\cos A} \cdot \frac{1}{\sin A} = \sec A \cdot \text{cosec}\, A$$
Hmm — re-checking: $\dfrac{1}{\sin A \cos A} = \dfrac{2}{2\sin A \cos A}$...
Wait — the question asks to prove equal to $2\cosec A$, which requires a specific approach. Using Exercise 8.3, Q4(vii): $\dfrac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \tan\theta$.
Using that result:
$$\text{LHS} = \tan A + \frac{\cos A}{\sin A} = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$$
$$= \frac{2}{2\sin A \cos A} = \frac{2}{\sin 2A}$$
But to match RHS $= 2\cosec A$, note the question likely intends $+\dfrac{\cos A}{\sin A}$ to be $\times \dfrac{\cos A}{\sin A}$ or the question is as given. Taking LHS as written and RHS $= 2\cosec A$:
Correct working:
$$\text{LHS} = \frac{\sin A(1-2\sin^2A)}{\cos A(2\cos^2A-1)}+\frac{\cos A}{\sin A} = \frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} = \frac{\sin^2A+\cos^2A}{\sin A\cos A} = \frac{1}{\sin A\cos A}$$
This equals $2\cosec A$ only if $\cos A = \tfrac{1}{2}$, so RHS should be $\sec A\cdot\cosec A$ or the question contains a typo. The standard textbook identity (Q4 vii) gives $\tan A$, so:
LHS $= \tan A + \cot A = \dfrac{\sin^2A+\cos^2A}{\sin A\cos A} = \dfrac{1}{\sin A\cos A} = \sec A\,\cosec A$ = RHS (if RHS is $\cosec A \sec A$).
Source: Chapter 8, Exercise 8.3, Q4(vii); Section 8.4 Trigonometric Identities
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The key steps examiners look for: (1) factorise numerator and denominator by taking $\sin A$ and $\cos A$ common respectively, (2) cancel the common factor $(1-2\sin^2A) = (2\cos^2A-1)$ using $\sin^2A+\cos^2A=1$, (3) add $\tan A + \cot A$ over a common denominator to get $\tfrac{1}{\sin A\cos A}$. Note: the printed question likely has RHS as $\sec A\,\cosec A$ (a common textbook variant); if your paper states $2\cosec A$, verify — it may be a misprint, as the algebra yields $\sec A\,\cosec A$.