$\,1(i).\,$ Prove that $\,\,\tan \theta+2\tan2\theta+4\cot4\theta=\cot\theta$
Sol. $\,\,\tan \theta+2\tan2\theta+4\cot4\theta\\=\tan\theta+2\tan2\theta+4.\frac{1-\tan^22\theta}{2\tan2\theta}\\=\tan\theta+2\tan2\theta+2(\cot2\theta-\tan2\theta)\\=\tan\theta+2\tan2\theta+2\cot2\theta-2\tan2\theta\\=\tan\theta+\frac{2(1-\tan^2\theta)}{2\tan\theta}\\=\tan\theta+\cot\theta-\tan\theta\\=\cot\theta$
$\,1(ii).\,$ Prove that $\,\,\cos^2(\theta+\phi)-\sin^2(\theta-\phi)=\cos2\theta \cos2\phi$
Sol. $\,\,\cos^2(\theta+\phi)-\sin^2(\theta-\phi)\\=\frac{1+\cos2(\theta+\phi)}{2}-\frac{1-\cos2(\theta-\phi)}{2}\\=\frac{\cos2(\theta+\phi)+\cos2(\theta-\phi)}{2}\\=\frac 12 \left[2\cos\frac{2(\theta+\phi)+2(\theta-\phi)}{2}\cos\frac{2(\theta+\phi)-2(\theta-\phi)}{2}\right]\\=\cos2\theta\cos2\phi$
$\,1(iii).\,$ Prove that $\,\,\frac{1+\cos2\alpha+\sin2\alpha}{1-\cos2\alpha+\sin2\alpha}=\cot \alpha$
Sol. $\,\,\frac{1+\cos2\alpha+\sin2\alpha}{1-\cos2\alpha+\sin2\alpha}\\=\frac{2\cos^2\alpha+2\sin\alpha\cos\alpha}{2\sin^2\alpha+2\sin\alpha\cos\alpha}\\=\frac{2\cos\alpha(\cos\alpha+\sin\alpha)}{2\sin\alpha(\sin\alpha+\cos\alpha)}\\=\cot\alpha$
$\,1(iv).\,$ Prove that $\,\,\cos^6A+\sin^6A=\frac 14(1+3\cos^22A)$
Sol. $\,\,\cos^6A+\sin^6A\\=(\cos^2A)^3+(\sin^2A)^3\\=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A\\+\sin^4A)\\=\cos^4A-\cos^2A\sin^2A+\sin^4A \\=\cos^4A+\sin^4A-\cos^2A\sin^2A\\=(\cos^2A)^2+(\sin^2A)^2-\cos^2A\sin^2A\\=(\cos^2A+\sin^2A)^2-2\sin^2A\cos^2A\\-\sin^2A\cos^2A\\=1-3\sin^2A\cos^2A\\=1-\frac 34(2\sin A\cos A)^2\\=1-\frac 34\sin^22A\\=1-\frac 34(1-\cos^22A)\\=\frac 14(1+3\cos^22A)$
$\,1(v).\,$ Prove that $\,\,\frac{1+\sin2\theta-\cos2\theta}{\sin\theta+\cos\theta}=2\sin \theta$
Sol. $\,\,\frac{1+\sin2\theta-\cos2\theta}{\sin\theta+\cos\theta}\\=\frac{1+2\sin\theta\cos\theta-1+2\sin^2\theta}{\sin\theta+\cos\theta}\\=\frac{2\sin\theta(\cos\theta+sin\theta)}{\sin\theta+\cos\theta}\\=2\sin\theta$
$\,1(vi).\,$ Prove that $\,\,\cot x\cos^2x-\tan x\sin^2x=2\cot 2x $
Sol. $\,\,\cot x\cos^2x-\tan x\sin^2x\\=\frac{\cos^3x}{\sin x}-\frac{\sin^3x}{\cos x}\\=\frac{\cos^4x-\sin^4x}{\sin x\cos x}\\=\frac{2(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)}{2\sin x\cos x}\\=\frac{2\cos 2x}{\sin 2x}\\=2\cot 2x$
$\,1(vii).\,$ Prove that $\,\,\frac{\sec 8\alpha-1}{\sec 4\alpha-1}=\frac{\tan 8\alpha}{\tan 2\alpha} $
Sol. $\,\,\frac{\sec 8\alpha-1}{\sec 4\alpha-1}\\=\frac{\frac{1-\cos8\alpha}{\cos 8\alpha}}{\frac{1-\cos4\alpha}{\cos4\alpha}}\\=\frac{\cos4\alpha}{\cos 8\alpha} \times \frac{1-\cos8\alpha}{1-\cos4\alpha}\\=\frac{\cos4\alpha}{\cos8\alpha} \times \frac{2\sin^24\alpha}{2\sin^22\alpha}\\=\frac{2\sin4\alpha\cos4\alpha\sin4\alpha}{2\sin^22\alpha\cos8\alpha}\\=\frac{\sin8\alpha.2\sin2\alpha\cos2\alpha}{2\sin^22\alpha\cos8\alpha}\\=\frac{\frac{\sin8\alpha}{\cos8\alpha}}{\frac{\sin2\alpha}{\cos2\alpha}}\\=\frac{\tan8\alpha}{\tan2\alpha}$
$\,1(viii).\,$ Prove that $\,\,\cos 4x-\cos 4y=8(\cos x-\cos y) \\ \times(\cos x+\cos y)(\cos x-\sin y)(\cos x+\sin y)$
Sol. $\,\,\cos 4x-\cos 4y\\=2\cos^22x-1-2\cos^22y+1\\=2(\cos^22x-\cos^22y)\\=2(\cos 2x+\cos 2y)(\cos 2x-\cos 2y)\\=2(2\cos^2x-1+1-2\sin^2y)\\ \times(2\cos^2x-1-2\cos^2y +1)\\=8(\cos^2x-\sin^2y)(\cos^2x-\cos^2y)\\=8(\cos x+\sin y)(\cos x-\sin y) \\ \times (\cos x+\cos y)(\cos x-\cos y)\\=8(\cos x-\cos y)(\cos x+\cos y) \\ \times(\cos x-\sin y)(\cos x+\sin y)$
$\,1(ix).\,$ Prove that $\,\,\frac{1+\cos A+\cos 2A}{\sin A+\sin 2A}=\cot A$
Sol. $\,\frac{1+\cos A+\cos 2A}{\sin A+\sin 2A}\\=\frac{1+\cos A+2\cos^2A-1}{\sin A+2\sin A\cos A}\\=\frac{\cos A(1+2\cos A)}{\sin A(1+2\cos A)}\\=\frac{\cos A}{\sin A}\\=\cot A$
$\,1(x).\,$ Prove that $\,\,\frac{\tan5A+\tan3A}{\tan5A-\tan3A}=4\cos4A\cos2A$
Sol. $\,\,\frac{\tan5A+\tan3A}{\tan5A-\tan3A}\\=\frac{\frac{\sin5A}{\cos5A}+\frac{\sin3A}{\cos3A}}{\frac{\sin5A}{\cos5A}-\frac{\sin3A}{\cos3A}}\\=\frac{\sin5A\cos3A+\cos5A\sin3A}{\sin5A\cos3A-\cos5A\sin3A}\\=\frac{\sin(5A+3A)}{\sin(5A-3A)}\\=\frac{\sin8A}{\sin2A}\\=\frac{2\sin4A\cos4A}{\sin2A}\\=\frac{2(2\sin2A\cos2A)\cos4A}{\sin2A}\\=4\cos4A\cos2A$
$\,1(xi).\,$ Prove that $\,\,\frac{\cos30^{\circ}-\sin20^{\circ}}{\cos40^{\circ}+\cos20^{\circ}}=\frac{4}{\sqrt3}\cos40^{\circ}\cos80^{\circ}$
Sol. $\,\,\frac{\cos30^{\circ}-\sin20^{\circ}}{\cos40^{\circ}+\cos20^{\circ}}\\=\frac{\sin60^{\circ}-\sin20^{\circ}}{\cos40^{\circ}+\cos20^{\circ}}\\=\frac{2\cos\frac{60^{\circ}+20^{\circ}}{2}\sin\frac{60^{\circ}-20^{\circ}}{2}}{2\cos\frac{40^{\circ}+20^{\circ}}{2}\cos\frac{40^{\circ}-20^{\circ}}{2}}\\=\frac{\cos40^{\circ}\sin20^{\circ}}{\cos30^{\circ}\cos10^{\circ}}\\=\frac{\cos40^{\circ} \times 2\sin10^{\circ}\cos10^{\circ}}{\frac{\sqrt3}{2}\cos10^{\circ}}\\=\frac{4}{\sqrt3}\cos40^{\circ}\sin10^{\circ}\\=\frac{4}{\sqrt3}\cos40^{\circ}\cos80^{\circ}$
Note : $\,\,\cos30^{\circ}=\sin(90^{\circ}-30^{\circ})=\sin60^{\circ}$
$\,1(xii).\,$ Prove that $\,\,4(\cos^310^{\circ}+\sin^320^{\circ})=3(\cos10^{\circ}+\sin20^{\circ})$
Sol. $\,\,4(\cos^310^{\circ}+\sin^320^{\circ})\\=4\cos^310^{\circ}+4\sin^320^{\circ}\\=\cos(3\times 10^{\circ})+3\cos10^{\circ}\\+3\sin20^{\circ}-\sin(3 \times 20^{\circ})\\=\cos30^{\circ}+3(\cos10^{\circ}+\sin20^{\circ})-\sin60^{\circ}\\=\frac{\sqrt3}{2}+3(\cos10^{\circ}+\sin20^{\circ})-\frac{\sqrt3}{2}\\=3(\cos10^{\circ}+\sin20^{\circ})$
Note : $\cos3\theta=4\cos^3\theta-3\cos\theta,\\ \sin3\theta=3\sin\theta-4\sin^3\theta$
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