(a) On the common domain of both sides, combine the fractions: \[\frac{\sin\theta(\sin\theta-\cos\theta)+\cos\theta(\sin\theta+\cos\theta)}{\sin^2\theta-\cos^2\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta-\cos^2\theta}.\] Dividing numerator and denominator by \(\cos^2\theta\) gives \((\tan^2\theta+1)/(\tan^2\theta-1)\), as required.

(b) \[\frac{\tan^2\theta+1}{\tan^2\theta-1}=2\quad\Longrightarrow\quad\tan^2\theta=3.\] The exact solutions are \[\theta=\frac\pi3,\quad\frac{2\pi}{3}.\] The point \(\theta=\pi/2\), where tangent is undefined, gives 1 in the original left-hand side and is not an additional solution.

(a) Multiplication by \(\cos\theta\) gives \(5\cos^2\theta-\sin^2\theta+\cos\theta=0\). Thus \(\boxed{6\cos^2\theta+\cos\theta-1=0}\), with \(\boxed{a=6,b=1,c=-1}\).

(b) \((3\cos\theta-1)(2\cos\theta+1)=0\). Thus \(\cos\theta=1/3\) or \(-1/2\), giving \(\boxed{\theta=1.23,\ 2.09,\ 4.19,\ 5.05}\) radians (3 s.f.). These all satisfy the original equation with \(\cos\theta\ne0\).

(a) Combining fractions gives \[\frac{\cos3x\cos x+\sin3x\sin x}{\sin x\cos x}=\frac{\cos(3x-x)}{\tfrac12\sin2x}=2\cot2x.\] This identity holds where the original denominators are nonzero.

(b) \(2\cot2x=4\) gives \(\tan2x=1/2\). Since \(0<2x<2\pi\), \[\boxed{x=\tfrac12\tan^{-1}(1/2)\approx0.232,\quad x=\tfrac12(\pi+\tan^{-1}(1/2))\approx1.80}.\] Neither is an excluded denominator zero.

Multiplying by \(\cos\theta\ne0\), \(\sin\theta(4\cos\theta+1)=0\). The roots of \(\sin\theta=0\) are excluded endpoints. Thus \(\cos\theta=-1/4\), giving \(\boxed{\theta=104.5^\circ}\) (1 d.p.).

(a) Combine over a common denominator: \[\frac{(1+\sin\theta)^2+\cos^2\theta}{\cos\theta(1+\sin\theta)}=\frac{2+2\sin\theta}{\cos\theta(1+\sin\theta)}=\frac2{\cos\theta},\] wherever the original expressions are defined.

(b) Thus \(2/\cos\theta=3/\sin\theta\), giving \(\tan\theta=3/2\). The two valid roots are \(\boxed{\theta\approx0.983,4.12\text{ radians}}\), from \(\tan^{-1}(3/2)\) and that angle plus \(\pi\).

Group the terms: \[3\cos\theta(2\tan\theta-1)+2(2\tan\theta-1)=0,\] \[(2\tan\theta-1)(3\cos\theta+2)=0.\] Thus \(\tan\theta=1/2\) or \(\cos\theta=-2/3\). The solutions in the stated interval are \[\boxed{\theta=26.6^\circ,\quad131.8^\circ}.\] The original expression is undefined at \(90^\circ\), which is not a solution.

(a) Replace \(\tan x\) by \(\sin x/\cos x\) and multiply numerator and denominator by \(\cos x\): \[\frac{\sin x+\sin x\cos x}{\sin x-\sin x\cos x}=\frac{\sin x(1+\cos x)}{\sin x(1-\cos x)}=\frac{1+\cos x}{1-\cos x}.\] Cancellation is valid wherever the original expression is defined.

(b) \[1+\cos x=k(1-\cos x),\quad(k+1)\cos x=k-1,\] so \[\boxed{\cos x=\frac{k-1}{k+1}}.\] The case \(k=-1\) has no solution.

(c) With \(k=4\), \(\cos x=3/5\). Thus \[\boxed{x=\pm\cos^{-1}(3/5)\approx\pm0.927\text{ radians}}.\] Both lie in the interval and satisfy the original expression with nonzero denominator.

(a) Where the expressions are defined, \[\frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}=\frac{(1+\sin x)^2-(1-\sin x)^2}{1-\sin^2x}\] \[=\frac{4\sin x}{\cos^2x}=\frac{4\tan x}{\cos x},\] proving the identity.

(b) The endpoint \(x=\pi/2\) is undefined. For the remaining interval, multiply by \(\cos x\): \[4\tan x=8\tan x\cos x,\qquad4\tan x(1-2\cos x)=0.\] Therefore either \(\tan x=0\), giving \(x=0\), or \(\cos x=1/2\), giving \(x=\pi/3\). Thus \[\boxed{x=0,\quad\pi/3}.\]

Clear the denominator: \[\tan\theta+2\sin\theta=3\tan\theta-6\sin\theta,\qquad2\tan\theta-8\sin\theta=0.\] For the defined part of the interval, multiply by \(\cos\theta\): \[2\sin\theta(1-4\cos\theta)=0.\] The roots of \(\sin\theta=0\) are excluded endpoints. Thus \(\cos\theta=1/4\), giving \[\boxed{\theta=75.5^\circ}.\] This value satisfies the original equation; \(90^\circ\) is undefined.

(a) Wherever the original expressions are defined, \[\left(\frac{1-\sin x}{\cos x}\right)\left(\frac{1+\sin x}{\sin x}\right)=\frac{1-\sin^2x}{\cos x\sin x}=\frac{\cos^2x}{\cos x\sin x}=\frac{\cos x}{\sin x}=\frac1{\tan x}.\]

(b) Use the identity: \(1/\tan x=2\tan^2x\), hence \(\tan^3x=1/2\). The tangent is positive only in the first quadrant of the permitted interval, giving \(\boxed{x=\tan^{-1}(2^{-1/3})\approx38.4^\circ}\). The endpoints and \(90^\circ\) do not define the original equation.

(a) With a common denominator, \[\frac{\sin\theta(1+\sin\theta)-\sin\theta(1-\sin\theta)}{1-\sin^2\theta}=\frac{2\sin^2\theta}{\cos^2\theta}=2\tan^2\theta.\]

(b) Thus \(2\tan^2\theta=8\), so \(\tan\theta=\pm2\). The two solutions are \(\boxed{63.4^\circ,116.6^\circ}\).

(a) Combine the fractions: \[\frac{\tan\theta(1-\cos\theta)+\tan\theta(1+\cos\theta)}{1-\cos^2\theta}=\frac{2\tan\theta}{\sin^2\theta}=\frac2{\sin\theta\cos\theta}.\]

(b) Use the identity: \(2/(\sin\theta\cos\theta)=6\cos\theta/\sin\theta\). Multiplying by the non-zero \(\sin\theta\cos\theta\) gives \(2=6\cos^2\theta\). Therefore \(\cos\theta=\pm1/\sqrt3\), giving \(\boxed{54.7^\circ,125.3^\circ}\). Neither is excluded by the original denominators.

(a) Combine the fractions: \[\frac{\sin^3\theta(1+\sin\theta)-\sin^2\theta(\sin\theta-1)}{(\sin\theta-1)(1+\sin\theta)}=\frac{\sin^4\theta+\sin^2\theta}{\sin^2\theta-1}.\] Since \(\sin^2\theta-1=-\cos^2\theta\), this is \[-\frac{\sin^2\theta}{\cos^2\theta}(1+\sin^2\theta)=-\tan^2\theta(1+\sin^2\theta).\] The identity holds where the original denominators are nonzero.

(b) Using the identity, \[-\tan^2\theta(1+\sin^2\theta)=\tan^2\theta(1-\sin^2\theta).\] Thus \(2\tan^2\theta=0\), so \(\tan\theta=0\). The only value in the open interval is \[\theta=\pi.\] It satisfies the original equation.

(a) Combining the fractions gives numerator \[(\sin\theta+2\cos\theta)(\cos\theta+2\sin\theta)-(\sin\theta-2\cos\theta)(\cos\theta-2\sin\theta)=4(\sin^2\theta+\cos^2\theta)=4.\] The denominator is \[\cos^2\theta-4\sin^2\theta=5\cos^2\theta-4.\] This proves the identity wherever the original denominators are nonzero.

(b) \[\frac4{5\cos^2\theta-4}=5\quad\Longrightarrow\quad25\cos^2\theta=24.\] Thus \(\cos\theta=\pm\sqrt{24/25}\), giving \[\theta\approx11.5^\circ\text{ or }168.5^\circ.\] The denominator is \(4/5\) at these roots, so both satisfy the original equation.

(a) Clear the denominator, replace \(\tan x=\sin x/\cos x\), and multiply by \(\cos x\): \[\sin x+\cos^2x=k(\sin x-\cos^2x).\] Use \(\cos^2x=1-\sin^2x\) and collect terms: \[(k+1)\sin^2x+(k-1)\sin x-(k+1)=0,\] as required. These steps apply wherever the original expression is defined.

(b) With \(k=4\), \[5\sin^2x+3\sin x-5=0,\qquad \sin x=\frac{-3\pm\sqrt{109}}{10}.\] Reject the negative root, which is less than \(-1\). The positive root is \(0.7440306\ldots\), giving \[\boxed{x=48.1^\circ,\quad131.9^\circ}.\] Both lie in the interval and satisfy the original equation with nonzero denominator.

(a) For \(\cos\theta\ne0\), \[\frac{1-2\sin^2\theta}{1-\sin^2\theta}=\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}=1-\tan^2\theta,\] proving the identity.

(b) Let \(u=\tan^2\theta\). The equation becomes \[1-u=2u^2,\qquad(2u-1)(u+1)=0.\] Reject \(u=-1\); hence \(\tan\theta=\pm1/\sqrt2\). The interval solutions are \[\boxed{\theta=35.3^\circ,\quad144.7^\circ}.\] Both satisfy the original equation; \(90^\circ\) is excluded because it is undefined.

(a) Multiplication gives \(8x^3+4x^2-2x-4x^2-2x+1=\boxed{8x^3-4x+1}\).

(b) Dividing numerator and denominator by \(\sec^2\theta=1+\tan^2\theta\), the ratio is \(1/(\sin^2\theta-\cos^2\theta)=\boxed{1/(1-2\cos^2\theta)}\), wherever the original expressions are defined.

(c) Put \(c=\cos\theta\). Then \(1=4c(1-2c^2)\), so \((2c-1)(4c^2+2c-1)=0\). Thus \(c=1/2,(-1+\sqrt5)/4,(-1-\sqrt5)/4\), giving \(\boxed{\theta=60^\circ,72^\circ,144^\circ}\). Each is in the required interval and has defined denominators.

(a) Multiplying by \(\sin x\), \(4\sin^2x+5\cos x+2=0\). Using \(\sin^2x=1-\cos^2x\) gives \(\boxed{4\cos^2x-5\cos x-6=0}\), with \(a=4,b=-5,c=-6\).

(b) \((4\cos x+3)(\cos x-2)=0\). Reject \(\cos x=2\); hence \(\cos x=-3/4\), giving \(\boxed{x=138.6^\circ,\ 221.4^\circ}\) (1 d.p.). Both have defined nonzero denominators in the original equation.

(a) Multiplying by \(\cos^2x\), \(3\sin^2x-3\sin^2x\cos^2x-4\cos^2x=0\). Substituting \(\sin^2x=1-\cos^2x\) gives \(\boxed{3\cos^4x-10\cos^2x+3=0}\), with \(\boxed{a=3,b=-10,c=3}\).

(b) \((3\cos^2x-1)(\cos^2x-3)=0\). Reject \(\cos^2x=3\); thus \(\cos x=\pm1/\sqrt3\), giving \(\boxed{x=54.7^\circ,125.3^\circ}\) (1 d.p.). Both have \(\cos x\ne0\), as required.

(a)(i) Expansion gives \(1+2\sin\theta\cos\theta=1\), hence \(\sin2\theta=0\). Thus \(\boxed{\theta=0,\pi/2,\pi}\). (ii) At \(0,\pi/2\) the unsquared sum is \(1\); at \(\pi\) it is \(-1\). Thus precisely \(\boxed{0,\pi/2}\) remain.

(b) With \(s=\sin\theta,c=\cos\theta\), the common denominator is \((c+s)(c-s)=c^2-s^2=1-2s^2\). The numerator is \(s(c-s)+(1-c)(c+s)=c+s-(c^2+s^2)=c+s-1\). This proves the identity where the denominators are nonzero.

(c) Write \(N=c+s-1\). Then \(N/(1-2s^2)=2N\), so \(N=0\) or \(1=2(1-2s^2)\). The first gives \(0,\pi/2\); the second gives \(\sin^2\theta=1/4\), hence \(\pi/6,5\pi/6\). All four have defined original denominators. Therefore \(\boxed{\theta=0,\pi/6,\pi/2,5\pi/6}\).

The principal value \(\alpha=\cos^{-1}(8/17)\) is acute, so \(\sin\alpha>0\). Therefore \[\sin\alpha=\sqrt{1-(8/17)^2}=\frac{15}{17},\qquad\tan\alpha=\frac{15}{8}.\] Hence \[\frac1{\sin\alpha}+\frac1{\tan\alpha}=\frac{17}{15}+\frac8{15}=\frac53.\]

(a) Start with the left-hand side (LHS):

\(\frac{\tan \theta + 7}{\tan^2 \theta - 3}\)

Use \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\):

\(\frac{\frac{\sin \theta}{\cos \theta} + 7}{\frac{\sin^2 \theta}{\cos^2 \theta} - 3} = \frac{\sin \theta\cos \theta + 7 \cos^2 \theta}{\sin^2 \theta - 3 \cos^2 \theta}\)

Use \(\sin^2 \theta + \cos^2 \theta = 1\) to rewrite the denominator:

\(\sin^2 \theta - 3 \cos^2 \theta = (1 - \cos^2 \theta) - 3 \cos^2 \theta = 1 - 4 \cos^2 \theta\)

Thus, the LHS becomes:

\(\frac{\sin \theta \cos \theta + 7 \cos^2 \theta}{1 - 4 \cos^2 \theta}\)

This matches the right-hand side (RHS), proving the identity.

(b) Using the identity from part (a), solve:

\(\frac{\tan \theta + 7}{\tan^2 \theta - 3} = \frac{5}{\tan \theta}\)

Cross-multiply to eliminate fractions:

\(\tan \theta (\tan \theta + 7) = 5 (\tan^2 \theta - 3)\)

\(\tan^2 \theta + 7 \tan \theta = 5 \tan^2 \theta - 15\)

Rearrange to form a quadratic equation:

\(4 \tan^2 \theta - 7 \tan \theta - 15 = 0\)

Factorize:

\((4 \tan \theta + 5)(\tan \theta - 3) = 0\)

Solutions for \(\tan \theta\):

\(\tan \theta = -1.25 \quad \text{and} \quad \tan \theta = 3\)

Find \(\theta\) in the range \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 71.6^\circ\) and \(\theta = 128.7^\circ\).

(a) Wherever defined, \[\frac1{1-\sin\theta}+\frac1{1+\sin\theta}=\frac{(1+\sin\theta)+(1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}=\frac2{1-\sin^2\theta}=\frac2{\cos^2\theta}.\]

(b) By (a), \(6/\cos^2\theta=12\), so \(\cos^2\theta=1/2\) and \(\cos\theta=\pm1/\sqrt2\). Thus \[\boxed{\theta=45^\circ,135^\circ,225^\circ,315^\circ}.\] All four satisfy the original denominators.

(a) For \(\cos\theta\ne0\), \[\frac1{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta}=\left(\frac{\sin\theta}{\cos\theta}\right)\sin\theta=\tan\theta\sin\theta.\]

(b) Using (a), \[2\tan\theta\sin\theta=\tan\theta\iff\tan\theta(2\sin\theta-1)=0.\] If \(\tan\theta=0\), \(\theta=0^\circ,180^\circ,360^\circ\). If \(\sin\theta=1/2\), \(\theta=30^\circ,150^\circ\). All have non-zero cosine, so \[\boxed{\theta=0^\circ,30^\circ,150^\circ,180^\circ,360^\circ}.\] Dividing by \(\tan\theta\) would lose three roots.

(a) Wherever both sides are defined, combine the fractions: \[\begin{aligned}\frac{\cos\theta}{1-\sin\theta}-\frac1{\cos\theta}&=\frac{\cos^2\theta-(1-\sin\theta)}{(1-\sin\theta)\cos\theta}\\&=\frac{1-\sin^2\theta-1+\sin\theta}{(1-\sin\theta)\cos\theta}\\&=\frac{\sin\theta(1-\sin\theta)}{(1-\sin\theta)\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta.\end{aligned}\]

(b) Using (a), \(2\tan\theta+3=0\), so \(\tan\theta=-3/2\). The reference angle is \(\tan^{-1}(3/2)=56.3099\ldots^\circ\). Tangent is negative in quadrants II and IV: \[\boxed{\theta=123.7^\circ,303.7^\circ}\quad(1\text{ d.p.}).\] Both satisfy the original denominator restrictions.

(a) Where \(\cos\theta\ne0\), \[\tan^4\theta-1=\frac{\sin^4\theta-\cos^4\theta}{\cos^4\theta}=\frac{(\sin^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta)}{\cos^4\theta}=\frac{1-2\cos^2\theta}{\cos^4\theta}.\]

(b) \[\frac{1-2\cos^2\theta}{\cos^2\theta}=7\Rightarrow\cos^2\theta=\frac19.\] Thus \(\theta=70.5^\circ,109.5^\circ\). The expression is undefined at \(90^\circ\).

(a) At points where the original expression is defined, \(1/\tan\theta=\cos\theta/\sin\theta\). Multiplying by \(\sin\theta\) gives \[6\sin^2\theta+\cos\theta=4.\] Using \(\sin^2\theta=1-\cos^2\theta\) and rearranging gives \(6\cos^2\theta-\cos\theta-2=0\).

(b) \[(3\cos\theta-2)(2\cos\theta+1)=0\Rightarrow\cos\theta=2/3,-1/2.\] Hence \[\theta=48.2^\circ,120^\circ,240^\circ,311.8^\circ.\] All four values leave the original denominators defined and nonzero.

(a) For \(\cos\theta\ne0\), \[3\tan^2\theta+5\sin^2\theta=\frac{3\sin^2\theta}{1-\sin^2\theta}+5\sin^2\theta=\frac{8\sin^2\theta-5\sin^4\theta}{1-\sin^2\theta}.\]

(b) Put \(u=\sin^2\theta\). Then \[8u-5u^2=9(1-u)\Rightarrow5u^2-17u+9=0\Rightarrow u=\frac{17\pm\sqrt{109}}{10}.\] Reject the value greater than 1. With \(u=(17-\sqrt{109})/10\), the acute angle is \(54.1^\circ\). All permitted angles are \(54.1^\circ,125.9^\circ,234.1^\circ\).

(a) Since sin²x=1−cos²x, the numerator is \(-\cos^2x-\cos x=-\cos x(1+\cos x)\). Division gives −cos x wherever the original expression is defined. (b) \(-\cos x/2=1/4\), so cos x=−1/2. Thus x=120°,240°. The undefined value x=180° is excluded.

(a) Expand the numerator: sin²θ+2 sin θ cos θ+cos²θ−1=2 sin θ cos θ. Divide by cos²θ to get 2 tan θ on the original domain. (b) \(2\tan\theta=5\tan^3\theta\), so \(\tan\theta(5\tan^2\theta-2)=0\). Thus θ=0° or \(\theta=\pm\tan^{-1}\sqrt{2/5}=\pm32.3^\circ\).

(a) \(\sin^2\theta=\cos\theta\), giving \(\cos^2\theta+\cos\theta-1=0\). Thus \(\cos\theta=(-1\pm\sqrt5)/2\). Reject the root less than \(-1\); the valid positive cosine gives \(\boxed{\theta=51.8^\circ,308.2^\circ}\) (1 d.p.). Both have defined tangent.

(b) Wherever the original denominators are nonzero, \(\tan\theta/\sin\theta-\sin\theta/\tan\theta=1/\cos\theta-\cos\theta=(1-\cos^2\theta)/\cos\theta=\sin^2\theta/\cos\theta=\boxed{\tan\theta\sin\theta}\).