Use the addition formula and \(\tan135^\circ=-1\). Writing \(T=\tan\theta\), \[k=\frac{T-1}{1+T}.\] Thus \[k+kT=T-1\quad\Rightarrow\quad(k-1)T=-1-k,\] \[\boxed{\tan\theta=\frac{1+k}{1-k}}.\] This expression applies where \(\tan\theta\) is defined; \(k=1\) corresponds to an undefined \(\tan\theta\).

(a) Expand \(R\cos(\theta-\alpha)=R\cos\alpha\cos\theta+R\sin\alpha\sin\theta\). Therefore \[R\cos\alpha=5,\qquad R\sin\alpha=3,\] giving \[\boxed{R=\sqrt{34}},\qquad\alpha=\tan^{-1}(3/5)=30.96375\ldots^\circ=\boxed{30.96^\circ}.\]

(b) \[\cos(\theta-\alpha)=\frac2{\sqrt{34}}.\] Let \(\delta=\cos^{-1}(2/\sqrt{34})\). The two solutions in the required interval are \(\theta=\alpha+\delta\) and \(\theta=\alpha-\delta\). Using unrounded values gives \[\boxed{\theta=100.9^\circ,\ -39.0^\circ}\quad\text{(1 d.p.)}.\]

Use \(\operatorname{cosec}^2\theta=1+\cot^2\theta\):

\[3(1+\cot^2\theta)=7+\cot^2\theta\ \Longrightarrow\ \cot^2\theta=2\ \Longrightarrow\ \sin^2\theta=\frac13.\]

The acute reference angle is \(\alpha=\sin^{-1}(1/\sqrt3)=35.264389\ldots^\circ\). Both signs of sine are allowed. The four solutions in the given open interval are

\[\boxed{\theta=-144.7^\circ,\ -35.3^\circ,\ 35.3^\circ,\ 144.7^\circ}.\]

Put \(t=\tan\theta>0\). The addition formula gives \[\frac1t\frac{t+1}{1-t}=7,\] where \(t=1\) is excluded because the original expression is undefined. Rearranging, \[7t^2-6t+1=0,\qquad t=\frac{3\pm\sqrt2}{7}.\] Both values are positive, giving \[\theta=12.8^\circ\quad\text{or}\quad32.2^\circ.\]

Use \(\tan^2\theta=\sec^2\theta-1\): \[2\sec^2\theta+3\sec\theta-20=0,\qquad(2\sec\theta-5)(\sec\theta+4)=0.\] Hence \(\cos\theta=2/5\) or \(\cos\theta=-1/4\). All four solutions in the given interval are \[\theta=-104.5^\circ,-66.4^\circ,66.4^\circ,104.5^\circ.\]

(a) Use the double-angle identities: \[\begin{aligned}\sin^22x+4\cos^2x\cos2x&=4\sin^2x\cos^2x+4\cos^2x(\cos^2x-\sin^2x)\\&=4\cos^4x.\end{aligned}\]

(b) The left side is \(4\cos^4x+5\), whose range is \([5,9]\). There are no real solutions precisely when \[k<5\quad\text{or}\quad k>9.\]

(c) Substitute \(x=t/2\) in the identity. The expression under the square root is \(4\cos^4(t/2)\). Its nonnegative square root is \(2\cos^2(t/2)=1+\cos t\). Therefore \[\int_{-\pi/3}^{\pi/3}(1+\cos t)\,dt=\left[t+\sin t\right]_{-\pi/3}^{\pi/3}=\frac{2\pi}{3}+\sqrt3.\]

(a) Expand and use the double-angle identities: \[\begin{aligned}4\cos\theta\sin(\theta+30^\circ)&=2\sqrt3\cos\theta\sin\theta+2\cos^2\theta\\&=\sqrt3\sin2\theta+\cos2\theta+1.\end{aligned}\] Matching \(R\cos(2\theta-\alpha)=R\cos\alpha\cos2\theta+R\sin\alpha\sin2\theta\), we obtain \(R\cos\alpha=1\), \(R\sin\alpha=\sqrt3\). Thus \[R=2,\qquad\alpha=60^\circ,\qquad k=1,\] and the required form is \(2\cos(2\theta-60^\circ)+1\).

(b) Apply part (a) with \(\theta=2\phi\), noting the coefficient is multiplied by 3: \[3\bigl(2\cos(4\phi-60^\circ)+1\bigr)=5\Rightarrow\cos(4\phi-60^\circ)=\frac13.\] Since \(-60^\circ<4\phi-60^\circ<300^\circ\), the relevant cosine angles are \(70.5288\ldots^\circ\) and \(289.4712\ldots^\circ\). Therefore \[\phi=32.6^\circ\quad\text{or}\quad87.4^\circ.\]

(a) Matching coefficients, \(R\cos\alpha=6\) and \(R\sin\alpha=4\). Therefore \[R=\sqrt{6^2+4^2}=2\sqrt{13},\qquad\alpha=\arctan(2/3)=33.69^\circ\quad(2\text{ d.p.}).\]

(b) Using the unrounded value of \(\alpha\), \[\sin(\theta-\alpha)=-\frac5{2\sqrt{13}}.\] Let \(\delta=\arcsin(5/(2\sqrt{13}))\). The two solutions are \[\theta=180^\circ+\delta+\alpha=257.6^\circ,\qquad\theta=360^\circ-\delta+\alpha=349.8^\circ.\] No other angles lie in the required interval.

(c) Halving the expression in part (a), \[3\sin4\beta-2\cos4\beta=\sqrt{13}\sin(4\beta-\alpha).\] Thus the greatest possible value is \(13+15=28\), attained when \(\sin(4\beta-\alpha)=\pm1\). The earliest positive value has \(4\beta-\alpha=90^\circ\), giving \[\beta=\frac{90^\circ+\alpha}{4}=30.9^\circ.\] The alternative angle \(-90^\circ\) gives a negative \(\beta\); the next value \(270^\circ\) gives a larger positive one.

(a) Expanding both cosine sums gives \[\left(\frac{\sqrt3}{2}\cos\theta-\frac12\sin\theta\right)\left(\frac12\cos\theta-\frac{\sqrt3}{2}\sin\theta\right)=\frac{\sqrt3}{4}(\cos^2\theta+\sin^2\theta)-\sin\theta\cos\theta=\frac{\sqrt3}{4}-\frac12\sin2\theta.\]

(b) \(\sin4\alpha=\sqrt3/2-2/5=0.466025\ldots\). With \(0<4\alpha<360^\circ\), \[4\alpha=\arcsin(0.466025\ldots)\ \text{or}\ 180^\circ-\arcsin(0.466025\ldots).\] Hence \(\alpha=6.94^\circ,\ 38.1^\circ\).

(c) In (a), substitute \(\theta=-10^\circ\) and \(10^\circ\). The two products are \(\sqrt3/4+\frac12\sin20^\circ\) and \(\sqrt3/4-\frac12\sin20^\circ\), whose sum is \(\sqrt3/2\).

(a) Expand: \[4\sin\theta\left(\frac12\sin\theta+\frac{\sqrt3}{2}\cos\theta\right)=2\sin^2\theta+2\sqrt3\sin\theta\cos\theta=1-\cos2\theta+\sqrt3\sin2\theta.\] Here \(R\cos\alpha=\sqrt3,\ R\sin\alpha=1\), so \(R=2,\alpha=30^\circ\). The result is \(1+2\sin(2\theta-30^\circ)\).

(b) \(\sin(2\theta-30^\circ)=-3/5\). Let \(\beta=\arcsin(3/5)=36.8699^\circ\). The solution from \(2\theta-30^\circ=-\beta\) is negative; the next is \(2\theta-30^\circ=180^\circ+\beta\). Hence the smallest positive solution is \(\theta=123.4^\circ\).

(a) Put \(t=\tan\theta\), where the expressions are defined. Then \[3\tan2\theta+\tan(\theta+45^\circ)=\frac{6t}{1-t^2}+\frac{t+1}{1-t}=\frac{6t+(t+1)^2}{1-t^2}=\frac{t^2+8t+1}{1-t^2}.\]

(b) \(t^2+8t+1=4(1-t^2)\Rightarrow5t^2+8t-3=0\). Hence \[t=\frac{-4\pm\sqrt{31}}5,\] giving \(\theta=17.4^\circ,\ 117.6^\circ\) in the required interval. Both satisfy the original equation.

(a) Wherever the original expressions are defined, \[2\sin\theta\operatorname{cosec}2\theta=\frac{2\sin\theta}{2\sin\theta\cos\theta}=\frac1{\cos\theta}=\sec\theta.\]

(b) Put \(s=\sec\theta\). Then \(s^2-1+\frac72s=8\), so \[2s^2+7s-18=0\Rightarrow s=\frac{-7\pm\sqrt{193}}4.\] Both values have \(|s|\ge1\). Taking \(\theta=\pm\arccos(1/s)\) gives \(\boxed{\theta=\pm0.952,\ \pm1.76}\) radians, all in the stated interval and in the original domain.

(c) Using \(\sin x=2\sin(x/2)\cos(x/2)\), \[8\sin^2(x/2)\operatorname{cosec}^2x=2\sec^2(x/2).\] Hence the integral is \(4\tan(x/2)+C\), on intervals where the original integrand is defined.

(a) Where defined, \[\sin2\theta(a\cot\theta+b\tan\theta)=2\sin\theta\cos\theta\left(a\frac{\cos\theta}{\sin\theta}+b\frac{\sin\theta}{\cos\theta}\right)=2a\cos^2\theta+2b\sin^2\theta=a+b+(a-b)\cos2\theta.\]

(b) The expression simplifies to \(8+2\cos2\theta\). Integrating the simplified expression (equivalently taking the limit at the removable singularity), \[\left[8\theta+\sin2\theta\right]_{-\pi/12}^{\pi/6}=2\pi+\frac{\sqrt3}{2}+\frac12.\]

(c) Substitute \(\theta=\alpha/3,\ a=2,b=7\) into (a). Then \(9-5\cos(2\alpha/3)=11\), so \(\cos(2\alpha/3)=-2/5\). With \(-2\pi/3<2\alpha/3<2\pi/3\), \[\alpha=\pm\frac32\arccos(-2/5)=\pm2.97\text{ radians}.\]

Using \(\sec^2\theta=1+\tan^2\theta\), \[\tan^2\theta=17-1=16.\] Since \(\theta\) is in the first quadrant, \(\tan\theta=4\). Therefore \[\tan(\theta+\pi/4)=\frac{4+1}{1-4}=-\frac53.\]

(a) Using double-angle identities, \[f(\theta)=6\sin2\theta+8\cos2\theta+8.\] Comparing with \[R\cos(2\theta-\alpha)=R\cos\alpha\cos2\theta+R\sin\alpha\sin2\theta\] gives \[R=\sqrt{8^2+6^2}=10,\quad k=8,\quad\tan\alpha=\frac68=\frac34.\] Thus \(\alpha=\tan^{-1}(3/4)\approx0.6435\) radians, and \[f(\theta)=10\cos(2\theta-\alpha)+8.\]

(b) \[\cos(2\theta-\alpha)=\frac{17-8}{10}=0.9.\] Let \(\beta=\cos^{-1}(0.9)\). The two first positive roots are \((\alpha-\beta)/2\) and \((\alpha+\beta)/2\); the smaller is \[\theta=\frac{\tan^{-1}(3/4)-\cos^{-1}(0.9)}2\approx0.0962\text{ radians}.\]

(c) \[\int f(\theta)\,d\theta=-3\cos2\theta+4\sin2\theta+8\theta+C.\] Equivalently, \(5\sin(2\theta-\alpha)+8\theta+C\), with exact \(\alpha=\tan^{-1}(3/4)\).

Using reciprocal identities, \[\frac1{\cos\theta}=\frac5{\sin\theta}.\] Both denominators must be non-zero. Multiplication gives \(\sin\theta=5\cos\theta\), so \(\tan\theta=5\). Tangent is positive in quadrants I and III. Therefore \[\theta\approx78.7^\circ,\quad258.7^\circ.\] Neither excluded denominator vanishes at these angles.

(a) Write the trigonometric ratios in terms of sine and cosine, and multiply by \(\sin\theta\cos\theta\): \[7\sin^2\theta+4\cos^2\theta-13\sin\theta=0.\] Using \(\cos^2\theta=1-\sin^2\theta\) gives \[3\sin^2\theta-13\sin\theta+4=0.\]

(b) Factorising, \[(3\sin\theta-1)(\sin\theta-4)=0.\] The value \(\sin\theta=4\) is impossible. Thus \(\sin\theta=1/3\), and \[\theta\approx19.5^\circ,\quad160.5^\circ.\] Both sine and cosine are non-zero at these angles, so both satisfy the original equation.

(a) Using \(\sin2\theta=2\sin\theta\cos\theta\), the first term is \(6\sin\theta\) wherever the original secant is defined. Expanding the cosine, \[10\cos(\theta-30^\circ)=5\sqrt3\cos\theta+5\sin\theta.\] The expression is therefore \[11\sin\theta+5\sqrt3\cos\theta=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha.\] Hence \[R=\sqrt{11^2+(5\sqrt3)^2}=14,\qquad\tan\alpha=\frac{5\sqrt3}{11},\qquad\alpha\approx38.21^\circ.\] The required form is \(14\sin(\theta+\alpha)\), using the exact angle for calculations.

(b) Replace \(\theta\) by \(2\beta\): \[14\sin(2\beta+\alpha)=2,\qquad\sin(2\beta+\alpha)=1/7.\] Since \(\alpha<2\beta+\alpha<180^\circ+\alpha\), the only admissible angle is \[2\beta+\alpha=180^\circ-\sin^{-1}(1/7).\] Therefore \[\boxed{\beta=\frac{180^\circ-\sin^{-1}(1/7)-\alpha}{2}\approx66.8^\circ}.\] This is not the excluded secant point \(\beta=45^\circ\).

(a) Wherever the original cotangent is defined, \[\sin2\theta\cot\theta-\cos2\theta=(2\sin\theta\cos\theta)\frac{\cos\theta}{\sin\theta}-(\cos^2\theta-\sin^2\theta)=\cos^2\theta+\sin^2\theta=1.\]

(b) Substitute \(\theta=\pi/12\) into the identity: \[\sin(\pi/6)\cot(\pi/12)=1+\cos(\pi/6)=\boxed{1+\frac{\sqrt3}2}.\]

(c) From part (a), \(\sin2\theta\cot\theta=1+\cos2\theta\). The equation becomes \[1-2\cos2\theta=1\quad\Longrightarrow\quad\cos2\theta=0.\] The smallest positive solution is \[\boxed{\theta=\pi/4}.\]

(a) The first quotient term is \(2x^2\); subtracting \(2x^2(2x+3)\) leaves \(10x^2+9x-15\). The next term is \(5x\); subtracting \(5x(2x+3)\) leaves \(-6x-15\). The next term is \(-3\), leaving \(-6\). Thus \[p(x)=(2x+3)(2x^2+5x-3)-6.\] The quotient is \(2x^2+5x-3\).

(b) \[\int\frac{p(x)}{2x+3}\,dx=\int\left(2x^2+5x-3-\frac6{2x+3}\right)\,dx=\boxed{\frac23x^3+\frac52x^2-3x-3\ln|2x+3|+C}.\]

(c) \[p(x)+6=(2x+3)(2x^2+5x-3)=(2x+3)(2x-1)(x+3).\] Thus \(\operatorname{cosec}2\theta=-3/2,\ 1/2,\ -3\). The middle value is impossible because it would give \(\sin2\theta=2\). The remaining values give \[\sin2\theta=-2/3\quad\text{or}\quad\sin2\theta=-1/3.\] Since \(0^\circ<2\theta<270^\circ\), only quadrant III roots are allowed: \[\theta=\frac{180^\circ+\sin^{-1}(1/3)}2\approx99.7^\circ,\qquad\theta=\frac{180^\circ+\sin^{-1}(2/3)}2\approx110.9^\circ.\]

(a) Use the compound-angle and double-angle formulae: \[\cos3\theta=\cos2\theta\cos\theta-\sin2\theta\sin\theta\] \[=(2\cos^2\theta-1)\cos\theta-2\sin^2\theta\cos\theta\] \[=2\cos^3\theta-\cos\theta-2(1-\cos^2\theta)\cos\theta=\boxed{4\cos^3\theta-3\cos\theta}.\]

(b) The expression equals \(\frac12(4\cos^3\theta-3\cos\theta)=\frac12\cos3\theta\), with \(\theta=5\pi/18\). Therefore \[\frac12\cos\frac{5\pi}6=\boxed{-\frac{\sqrt3}4}.\]

(c) From (a), \(4\cos^3x=\cos3x+3\cos x\) and \(4\cos^3 3x=\cos9x+3\cos3x\). Thus \[12\cos^3x-4\cos^3 3x=3\cos3x+9\cos x-\cos9x-3\cos3x=9\cos x-\cos9x.\] Integrate term by term: \[\boxed{9\sin x-\frac19\sin9x+C}.\]

(a) Expand the compound angle: \[4\sin x\left(\sin x\cos\frac\pi6+\cos x\sin\frac\pi6\right)=2\sqrt3\sin^2x+2\sin x\cos x.\] Now \(2\sin^2x=1-\cos2x\) and \(2\sin x\cos x=\sin2x\), giving the required identity.

(b) Integrate the right-hand side: \[\left[\sqrt3x-\frac{\sqrt3}2\sin2x-\frac12\cos2x\right]_0^{5\pi/6}.\] At the upper limit, \(\sin(5\pi/3)=-\sqrt3/2\) and \(\cos(5\pi/3)=1/2\). Thus \[\frac{5\sqrt3\pi}{6}+\frac34-\frac14+\frac12=\boxed{\frac{5\sqrt3\pi}{6}+1}.\]

(c) Put \(x=2y\) in (a): \[\sqrt3-\sqrt3\cos4y+\sin4y=\sqrt3,\qquad\sin4y=\sqrt3\cos4y.\] Cosine cannot be zero in this equation, so \(\tan4y=\sqrt3\). Hence \(4y=\pi/3+n\pi\). The smallest positive value occurs for n=0: \[\boxed{y=\pi/12}.\]

Expand both angles: \[\frac{\sqrt3}{2}\sin2\theta+\frac12\cos2\theta=5\left(\frac12\cos2\theta-\frac{\sqrt3}{2}\sin2\theta\right).\] Collect terms: \[3\sqrt3\sin2\theta=2\cos2\theta.\] If \(\cos2\theta=0\), this equation would require \(\sin2\theta=0\) too, which is impossible. Hence \[\tan2\theta=\frac2{3\sqrt3}.\] In \(0^\circ<2\theta<360^\circ\), the values are \(2\theta=21.0517\ldots^\circ\) and \(201.0517\ldots^\circ\). Therefore \[\boxed{\theta=10.5^\circ,\quad100.5^\circ}.\]

First expand as required, using \(\cosec\theta=1/\sin\theta\): \[\frac{\sin\theta\cos30^\circ+\cos\theta\sin30^\circ}{\sin\theta}=2.\] Thus \[\frac{\sqrt3}{2}+\frac12\cot\theta=2,\qquad\cot\theta=4-\sqrt3,\qquad\tan\theta=\frac1{4-\sqrt3}.\] The reference angle is \(23.79398\ldots^\circ\). Tangent is positive in quadrants I and III, giving \[\boxed{\theta=23.8^\circ,\quad203.8^\circ}.\] Values with \(\sin\theta=0\) are excluded by the original cosecant expression.

Write the expression in sine and cosine, retaining the original restrictions \(\sin\theta\ne0\), \(\cos\theta\ne0\): \[\frac1{\cos^2\theta}\frac{\cos\theta}{\sin\theta}=\frac1{\sin\theta\cos\theta}=8.\] Thus \[\sin2\theta=1/4.\] In \(0<2\theta<2\pi\), sine is positive in the first two quadrants. Hence \[\theta=\tfrac12\arcsin(1/4),\quad\tfrac12(\pi-\arcsin(1/4)),\] \[\boxed{\theta=0.126,\quad1.44\text{ radians}}.\] Neither value violates the original restrictions.

(a) Expanding gives \(R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x\). Therefore \[R\cos\alpha=5\sqrt3,\qquad R\sin\alpha=5,\] \[R=\sqrt{75+25}=10,\qquad\tan\alpha=1/\sqrt3,\qquad\alpha=\pi/6.\] The required form is \(\boxed{10\cos(x-\pi/6)}\).

(b) The expression becomes \(4+10\cos(x-\pi/6)\). Its minimum is \(4-10=\boxed{-6}\), when the cosine is −1. In the stated interval, \(x-\pi/6=-\pi\), giving \(\boxed{x=-5\pi/6}\).

(c) Use (a) with \(x=3\theta\). The denominator is \(100\cos^2(3\theta-\pi/6)\), so \[\int\frac1{100}\sec^2(3\theta-\pi/6)\,d\theta=\boxed{\frac1{300}\tan(3\theta-\pi/6)+C}.\]