Exercise 5.3 1. Prove each of the trigonometric identities. \( \begin{array}{ll}\text { ea) } \tan ^{2} A \cdot \cos ^{2} A+\frac{\sin ^{2} A}{\tan ^{2} A}=1 & \text { b) }(\cos x+\sin x)(\cos x-\sin x)=2 \cos ^{2} x- \\ \text { e) } \sin ^{4} x-\cos ^{4} x=2 \sin ^{2} x-1 & \text { d) } \cos x \sin ^{2} x+\cos ^{3} x=\cos x \\ \text { e) } \tan ^{2} \theta-\sin ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta & \text { (f) } \frac{1+2 \sin x \cos x}{\sin ^{2} x-\cos ^{2} x}=\frac{\sin x+\cos x}{\sin x-\cos x} \\ \text { eg) }\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}=\frac{1+\sin \theta}{1-\sin \theta} & \text { h) } \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{2}{\sin x} \\ \text { i) } \frac{\sin A}{1-\cos A}=\frac{1}{\sin A}+\frac{1}{\tan A} & \begin{array}{ll}\text { j) } \frac{1}{\cos \theta}-\frac{\cos \theta}{1+\sin \theta}=\tan \theta & \text { cos } \\ \text { k) } \frac{\cos }{1-\cos x}+\frac{\cos x}{1+\cos x}=\frac{2 \cos x}{\sin ^{2} x} \\ \text { l) } \frac{\cos x}{\tan x}+\sin x=\frac{1}{\sin x}\end{array}\end{array}> \)

**a)** We start with \[ \tan^2 A\cdot \cos^2 A+\frac{\sin^2 A}{\tan^2 A}. \] Recall that \[ \tan^2 A=\frac{\sin^2 A}{\cos^2 A}. \] Thus, the first term becomes \[ \frac{\sin^2 A}{\cos^2 A}\cdot \cos^2 A=\sin^2 A. \] For the second term, substitute for \(\tan^2 A\): \[ \frac{\sin^2 A}{\tan^2 A}=\frac{\sin^2 A}{\frac{\sin^2 A}{\cos^2 A}}=\cos^2 A. \] Adding these gives \[ \sin^2 A+\cos^2 A=1. \] --- **b)** Examine the product \[ (\cos x+\sin x)(\cos x-\sin x). \] This is a difference of two squares: \[ \cos^2 x-\sin^2 x. \] But since \[ \cos^2 x-\sin^2 x=2\cos^2 x-1 \quad \text{(because } \sin^2 x=1-\cos^2 x\text{)}, \] the identity is verified. --- **e)** Show that \[ \sin^4 x-\cos^4 x=2\sin^2 x-1. \] Write the left-hand side as a difference of squares: \[ \sin^4 x-\cos^4 x=(\sin^2 x-\cos^2 x)(\sin^2 x+\cos^2 x). \] Since \[ \sin^2 x+\cos^2 x=1, \] we have \[ \sin^4 x-\cos^4 x=\sin^2 x-\cos^2 x. \] Now, express \(\sin^2 x-\cos^2 x\) in an alternative form. Note that \[ \cos^2 x=1-\sin^2 x, \] so \[ \sin^2 x-\cos^2 x=\sin^2 x-(1-\sin^2 x)=2\sin^2 x-1. \] --- **d)** Prove \[ \cos x\, \sin^2 x+\cos^3 x=\cos x. \] Factor out \(\cos x\): \[ \cos x\,\sin^2 x+\cos^3 x=\cos x\Bigl(\sin^2 x+\cos^2 x\Bigr). \] Since \[ \sin^2 x+\cos^2 x=1, \] this simplifies to \[ \cos x. \] --- **e)** (Second occurrence) Prove that \[ \tan^2 \theta-\sin^2 \theta=\tan^2 \theta\,\sin^2 \theta. \] Write \(\tan^2 \theta\) in terms of sine and cosine: \[ \tan^2 \theta=\frac{\sin^2 \theta}{\cos^2 \theta}. \] Then the left-hand side becomes \[ \frac{\sin^2 \theta}{\cos^2 \theta}-\sin^2 \theta=\sin^2 \theta\left(\frac{1}{\cos^2 \theta}-1\right). \] Express \(1- \cos^2 \theta\) in the numerator: \[ \frac{1}{\cos^2\theta}-1=\frac{1-\cos^2\theta}{\cos^2\theta}=\frac{\sin^2\theta}{\cos^2\theta}. \] Thus, \[ \tan^2 \theta-\sin^2 \theta=\sin^2 \theta\cdot\frac{\sin^2 \theta}{\cos^2\theta}=\