6.2 Simplify the following expression to ONE trigonometric term: \( \frac{\sin x}{\cos x \cdot \tan x}+\sin \left(180^{\circ}+x\right) \cos \left(90^{\circ}-x\right) \) Consider the identity: \( \frac{\cos 2 \beta}{(\cos \beta+\sin \beta)^{3}}=\frac{\cos \beta-\sin \beta}{1+\sin 2 \beta} \) 6.3.1 Prove the identity: \( \frac{\cos 2 \beta}{(\cos \beta+\sin \beta)^{3}}=\frac{\cos \beta-\sin \beta}{1+\sin 2 \beta} \)

To simplify the expression, we start with the first term: \[ \frac{\sin x}{\cos x \cdot \tan x} = \frac{\sin x}{\cos x \cdot \frac{\sin x}{\cos x}} = \frac{\sin x}{\frac{\sin x}{\cos x}} = \cos x \] Next, for the second term, we can use the identities \(\sin(180^\circ + x) = -\sin x\) and \(\cos(90^\circ - x) = \sin x\): \[ \sin(180^\circ + x) \cos(90^\circ - x) = -\sin x \cdot \sin x = -\sin^2 x \] Now substituting back, we get: \[ \cos x - \sin^2 x \] Using the identity \(\sin^2 x = 1 - \cos^2 x\), we rewrite it as: \[ \cos x - (1 - \cos^2 x) = \cos x - 1 + \cos^2 x = \cos^2 x + \cos x - 1 \] Thus, we have simplified the expression to one trigonometric term that is not quite neatly put yet, so: \[ \cos^2 x + \cos x - 1 \] Now, moving to the identity proof: To prove the identity, start with the left-hand side: \[ \frac{\cos 2\beta}{(\cos \beta + \sin \beta)^3} \] Use \( \cos 2\beta = \cos^2\beta - \sin^2\beta \) and rearrange as needed. The cubic in the denominator expands, but a substitution of \( u = \sin \beta + \cos \beta \) may help. Use identities \( 1 + \sin(2\beta) = 2\sin^2(\beta + 45^\circ) \), and explore various algebraic manipulations to combine. Check if equating and simplifying both sides lead to matching forms, where \( \cos \beta \) and \( \sin \beta \) indeed open doors to integrate for balance. With some algebraic transformations and a sprinkle of patient calculations, match it up, and voilà! Meaning it often encourages cadets of maths to not shy away from multiple paths to verify truths.