(b) Prove the following identities: (1) \( \tan x+\frac{1}{\tan x}=\frac{1}{\sin x \cos x} \) (3) \( \frac{1}{\tan x}-\tan x=\frac{1-2 \sin ^{2} x}{\sin x \cos x} \) (2) \( \frac{1}{\sin \alpha}-\sin \alpha=\frac{\cos \alpha}{\tan \alpha} \) (5) \( \frac{\tan ^{2} \theta}{\tan ^{2} \theta+1}=\sin ^{2} \theta \) (6) \( \frac{\tan \theta}{1-\tan ^{2} \theta}=\frac{\cos \theta \sin \theta}{\cos ^{2} \theta-\sin ^{2} \theta} \) (7) \( \frac{\cos x}{1-\tan x}+\frac{\sin x \tan x}{\tan x-1}=\sin x+\cos x \) (8) \( \tan ^{2} x\left(1+\frac{1}{\tan ^{2} x}\right)=\frac{1}{\cos ^{2} x} \) (9) \( \frac{1+\tan ^{2} x}{1+\frac{1}{\tan ^{2} x}}=\tan ^{2} x \) (10) \( \frac{\tan x+\sin x}{\tan x-\sin x}=\frac{(1+\cos x)^{2}}{\sin ^{2} x} \) 123

**(1)** We start with the left-hand side (LHS): \[ \tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}. \] Combine the two fractions over a common denominator: \[ \frac{\sin^2x+\cos^2x}{\sin x\cos x}. \] Since \[ \sin^2x+\cos^2x=1, \] the expression becomes \[ \frac{1}{\sin x\cos x}, \] which is the right-hand side (RHS). --- **(3)** The left-hand side is \[ \frac{1}{\tan x}-\tan x=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}. \] Combine these fractions: \[ \frac{\cos^2x-\sin^2x}{\sin x\cos x}. \] Recall that \[ \cos^2x-\sin^2x=1-2\sin^2x \] (since both \(\cos^2x-\sin^2x\) and \(1-2\sin^2x\) equal \(\cos2x\)). Hence, \[ \frac{\cos^2x-\sin^2x}{\sin x\cos x}=\frac{1-2\sin^2x}{\sin x\cos x}, \] which proves the identity. --- **(2)** The left-hand side is \[ \frac{1}{\sin \alpha}-\sin \alpha. \] Write over a common denominator: \[ \frac{1-\sin^2\alpha}{\sin\alpha}. \] Using the Pythagorean identity, we have \[ 1-\sin^2\alpha=\cos^2\alpha. \] Thus, \[ \frac{\cos^2\alpha}{\sin\alpha}. \] Now, the right-hand side is \[ \frac{\cos\alpha}{\tan\alpha}=\frac{\cos\alpha}{\frac{\sin\alpha}{\cos\alpha}}=\frac{\cos^2\alpha}{\sin\alpha}. \] Both sides are equal. --- **(5)** Rewrite the left-hand side using the identity \[ 1+\tan^2\theta=\sec^2\theta=\frac{1}{\cos^2\theta}: \] \[ \frac{\tan^2\theta}{\tan^2\theta+1}=\tan^2\theta\cos^2\theta. \] Since \[ \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}, \] we have \[ \tan^2\theta\cos^2\theta=\frac{\sin^2\theta}{\cos^2\theta}\cdot\cos^2\theta=\sin^2\theta. \] --- **(6)** Express \(\tan \theta\) in terms of sine and cosine: \[ \frac{\tan\theta}{1-\tan^2\theta}=\frac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\sin^2\theta}{\cos^2\theta}}. \] Combine the denominator: \[ 1-\frac{\sin^2\theta}{\cos^2\theta}=\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}. \] Thus, \[ \frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}=\frac{\sin\theta}{\cos\theta}\cdot\frac{\cos^2\theta}{\cos^2\theta-\sin^2\theta}=\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}. \] --- **(7)** Examine the expression: \[ \frac{\cos x}{1-\tan x}+\frac{\sin x\,\tan x}{\tan x-1}. \] Notice that \[ \tan x-1=-(1-\tan x). \] Thus, the second term can be rewritten as: \[ \frac{\sin x\,\tan x}{\tan x-1}=-\frac{\sin x\,\tan x}{1-\tan x}. \] Now the expression becomes: \[ \frac{\cos x}{1-\tan x}-\frac{\sin x\,\tan x}{1-\tan x}=\frac{\cos x-\sin x\,\tan x}{1-\tan x}. \] Express \(\tan x\) as \(\frac{\sin x}{\cos x}\): \[ \sin x\,\tan x=\frac{\sin^2 x}{\cos x}. \] Thus, the numerator becomes: \[ \cos x-\frac{\sin^2 x}{\cos x}=\frac{\cos^2 x-\sin^2 x}{\cos x}. \] The entire expression is then: \[ \frac{\frac{\cos^2 x-\sin^2 x}{\cos x}}{1-\tan x}. \] Recall that \[ 1-\tan x=\frac{\cos x-\sin x}{\cos x}, \] so we have: \[ \frac{\cos^2 x-\sin^2 x}{\cos x-\sin x}. \] Notice that \[ \cos^2 x-\sin^2 x=(\cos x-\sin x)(\cos x+\sin x). \] Cancel the common factor \((\cos x-\sin x)\): \[ \frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x-\sin x}=\cos x+\sin x. \] --- **(8)** Simplify the left-hand side: \[ \tan^2 x\left(1+\frac{1}{\tan^2 x}\right)=\tan^2x+\tan^2x\cdot\frac{1}{\tan^2x}=\tan^2x+1. \] Recall the Pythagorean identity for tangent and secant: \[ 1+\tan^2x=\sec^2x=\frac{1}{\cos^2x}. \] --- **(9)** Write the denominator with a common denominator: \[ 1+\frac{1}{\tan^2x}=\frac{\tan^2x+1}{\tan^2x}. \] Thus, the expression becomes: \[ \frac{1+\tan^2x}{\frac{\tan^2x+1}{\tan^2x}}=\tan^2x. \] --- **(10)** Rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\): For the numerator: \[ \tan x+\sin x=\frac{\sin x}{\cos x}+\sin x=\sin x\left(\frac{1}{\cos x}+1\right)=\frac{\sin x(1+\cos x)}{\cos x}. \] For the denominator: \[ \tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\sin x\left(\frac{1}{\cos x}-1\right)=\frac{\sin x(1-\cos x)}{\cos x}. \] Divide the numerator by the denominator: \[ \frac{\frac{\sin x(1+\cos x)}{\cos x}}{\frac{\sin x(1-\cos x)}{\cos x}}=\frac{1+\cos x}{1-\cos x}. \] Now, use the identity \[ 1-\cos^2x=\sin^2x, \] which can be rearranged as \[ 1-\cos x=\frac{\sin^2x}{1+\cos x}, \] provided \(1+\cos x\neq 0\). Substitute into the expression: \[ \frac{1+\cos x}{1-\cos x}=\frac{1+\cos x}{\frac{\sin^2x}{1+\cos x}}=\frac{(1+\cos x)^2}{\sin^2x}. \]