Question 12 Marks
Prove that: $\tan(-225^\circ)\cot(-405^\circ)-\tan(765^\circ)+\cot(675^\circ)=0$
Answer$\text{L.H.S}=\tan(-225^\circ)\cot(-405^\circ)-\tan(-765)\cot(675^\circ)$ $=-\tan225^\circ(-\cot405^\circ)+\tan765^\circ\cot765\circ$ $\begin{pmatrix}\because\tan(-\theta)=-\tan\theta\\\&\cot(-\theta)=-\cot\theta\end{pmatrix}$ $=\tan\Big(\pi+\frac{\pi}{4}\Big)\cot\Big(2\pi\frac{\pi}{4}\Big)+\tan\Big(4\pi+\frac{\pi}{4}\Big)\cot\Big(4\pi-\frac{\pi}{4}\Big)$ $=\tan\frac{\pi}{4}\cot\frac{\pi}{4}+\tan\frac{\pi}{4}\times\Big(-\cot\frac{\pi}{4}\Big)$ $(\because\cot(4\pi-\theta)=-\cot\theta)$ $= 1. 1 + 1 (-1) $ $= 1 - 1$ $= 0$ $=\text{R.H.S}$ $\text{Proved}$
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Find x From the following equation: $\text{ x}\cot\Big(\frac{\pi}{2}+\theta\Big)+\tan\Big(\frac{\pi}{2}\Big)\sin\theta=\text{coses}\Big(\frac{\pi}{2}+\theta\Big)=0$
AnswerWe have: $\text{ x}\cot(90^\circ+\theta)+\tan(90^\circ+\theta)\sin\theta=\text{coses}(90^\circ+\theta)=0$ $\Rightarrow\text{x}(-\tan\theta)-\cot\theta\times\sin\theta+\sec\theta=0$ $\Rightarrow\text{-x}\tan\theta-\frac{\cos\theta}{\sin\theta}\times\sin\theta+\frac{1}{\cos\theta}=0$ $\Rightarrow\text{-x}\frac{\sin\theta}{\cos\theta}-\cos\theta+\frac{1}{\cos\theta}=0$ $\Rightarrow\frac{\text{-x}\sin\theta-\cos^2\theta+1}{\cos\theta}=0$ $\Rightarrow\text{-x}\sin\theta+1-\cos^2\theta=0$ $\Rightarrow\text{-x}\sin\theta+\sin^2\theta=0$ $\Rightarrow\text{x}\sin\theta=\sin^2\theta$ $\Rightarrow\text{x}=\frac{\sin^2\theta}{\sin\theta}$ $\Rightarrow\text{x}=\sin\theta$
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Find the value of the following trigonomentric ratio: $\cos\Big(\frac{39\pi}{4}\Big)$
Answer$\cos1755^\circ=\cos\Big(10\pi-\frac{\pi}{4}\Big)$ $=\cos\Big(2\times5\pi-\frac{\pi}{4}\Big)$ $=\cos\frac{\pi}{4}$ $(\because\cos(2\text{k}\pi-\theta)=\cos\theta,\text{ k}\in\text{N})$ $=\frac{1}{\sqrt{2}}$
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If A, B, C, D be the angles of a cyclic quadrilateral, take in order, proved that: $\cos(180^\circ-\text{A})+\cos(180^\circ+\text{B})+\cos(180^\circ+\text{C})-\sin(90^\circ+\text{D})=0$
Answer$\because \text{A, B, C, D}$ are the angles of a cycle quardialateral in orcer, $\because\text{ A + C} = \pi\& \text{ B + D} =\pi $ $\Rightarrow\pi -\text{A = C}$ & $\pi- \text{D = B}$ $\Rightarrow\cos(\pi-\text{C})=\cos\text{C}\ \dots(\text{i})$ $$& $\cos(\pi-\text{D})=\cos\text{B}\dots(\text{ii})$ Now, $\cos(180^\circ-\text{A})+\cos(180^\circ+\text{B})+\cos(180^\circ+\text{C})-\sin(90^\circ+\text{D})$ $=\cos\text{C}+(-\cos\text{B})-\cos\text{C}-\cos\text{D}$ $(\because\cos(180^\circ-\text{B})=-\cos\text{B},\cos(180^\circ+\text{C})=-\cos\text{ C }\& \text{ using }( \text{i}))$ $=-\cos\text{B}-\cos\text{D}$ $=-\cos\text{B}-(-\cos\text{B})$ (Using (ii)) $=-\cos\text{}+\cos\text{B}$ $= 0$ $\text{Proved}$
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Prove that: $\sin\frac{10\pi}{3}\cos\frac{13\pi}{6}+\cos\frac{8\pi}{3}\sin\frac{5\pi}{6}=-1$
Answer$\text{L.H.S}=\sin\frac{10\pi}{3}\cos\frac{13\pi}{6}+\cos\frac{8\pi}{3}\sin\frac{5\pi}{6}$ $=\sin600^\circ\cos390^\circ+\cos480^\circ\sin150^\circ$ $=\sin\Big(3\pi+\frac{\pi}{3}\Big)\cos\Big(2\pi+\frac{\pi}{6}\Big)+\cos\Big(3\pi-\frac{\pi}{3}\Big)\sin\Big(\pi-\frac{\pi}{6}\Big)$ $=-\sin\frac{\pi}{3}\cos\frac{\pi}{6}-\cos\frac{\pi}{3}-\sin\frac{\pi}{6}$$\Big(\because\sin\Big(3\pi+\frac{\pi}{3}\Big)=-\sin\frac{\pi}{3 }\&\cos\Big(3\pi-\frac{\pi}{3}\Big)=-\cos\frac{\pi}{3}\Big)$ $=\frac{-\sqrt3}{2}\times\frac{-\sqrt3}{2}-\frac{1}{2}\times\frac{1}{2}$ $=\frac{-3}{4}-\frac{1}{4}$ $=\frac{-4}{4}$ $=-1$ $=\text{R.H.S}$ $\text{Proved}$
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Find the value of the following trigonomentric ratio: $\text{cosec}\Big(-\frac{20\pi}{3}\Big)$
Answer$\text{cosec}(-1200^\circ)=\text{cosec}\Big(-\Big(7\pi+\frac{\pi}{3}\Big)\Big)$ $=\text{cosec}\Big(7\pi-\frac{\pi}{3}\Big)$ $(\because\text{cosec}(-\theta)=-\text{cosec}\theta)$ $=-\text{cosec}\Big(2\times3\pi+\Big(\pi-\frac{\pi}{3}\Big)\Big)$ $=-\text{cosec}\big(\pi-\frac{\pi}{3}\big)$ $\begin{pmatrix}\because\text{cosec is period 2}\pi,\\\therefore\text{cosec}(2\pi+\theta)=2\text{n}\pi+\theta\\=\text{cosec}\theta\text{ for all n}\in\text{N}\end{pmatrix}$ $=-\text{cosec}\frac { \pi}{ 3}$ $(\therefore\text{cosec}(\pi-\theta)=\text{cosec})$ $=\frac{-2}{\sqrt{3}}$
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Find the value of the following trigonomentric ratio: $\sin\Big(\frac{17\pi}{6}\Big)$
Answer$\sin510^\circ=\sin\Big(3\pi-\frac{\pi}{6}\Big)$ $=\sin\frac{(\pi)}{6}$ $\Big(\because3\pi-\frac{\pi}{6}$ lies in second quadrant $\Big)$ $=\frac{1}{2}$ Alternative solution $\sin510^\circ=\sin\Big(3\pi-\frac{\pi}{6}\Big)$ $=\sin\bigg(2\pi+\Big(\pi-\frac{\pi}{6}\Big)\bigg)$ $\sin\Big(\pi-\frac{\pi}{6}\Big)$ $(\because\sin(2\pi+\theta)=\sin\theta,$ as sine is periodic with period $2\pi)$ $=\sin\Big(\frac{\pi}{6}\Big)$ $(\because \sin(\pi-\theta)=\sin\theta)$ $=\frac{1}{2}$
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Find x from the following equation: $\text{coses}\Big(\frac{\pi}{2}+\theta\Big)+\text{x}\cos\theta\cot\Big(\frac{\pi}{2}+\theta\Big)=\sin\Big(\frac{\pi}{2}+\theta\Big)$
Answer$\text{coses}(90^\circ+\theta)+\text{x}\cos\theta\cot(90^\circ+\theta)=\sin(90^\circ+\theta)$ $\Rightarrow \sec+\text{x}\cos\theta\times(-\tan\theta)=\cos\theta$ $\Rightarrow\frac{1}{\cos\theta}+\text{x}\cos\theta\times\frac{(-\sin\theta)}{\cos\theta}=\cos\theta$ $\Rightarrow\frac{1}{\cos\theta}-\text{x}\sin\theta=\cos\theta$ $\Rightarrow\frac{1-\text{x}\sin\theta\cos\theta}{\cos\theta}=\cos\theta$ $\Rightarrow1-\text{x}\sin\theta\cos\theta=\cos^2\theta$ $\Rightarrow1-\cos^2\theta=\text{x}\sin\theta\cos\theta$ $\Rightarrow\sin^2\theta=\text{x}\sin\theta\cos\theta$ $\Rightarrow\sin\theta=\text{x}\cos\theta$ $\Rightarrow\text{x}=\frac{\sin\theta}{\cos\theta}$ $=\tan\theta$ $\text{Hence x}=\tan\theta$
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Find the value of the following trigonomentric ratio: $\tan\frac{11\pi}{6}$
Answer$\tan\frac{11\pi}{6}=\tan\Big(2\pi-\frac{\pi}{6}\Big)$ $=-\tan\frac{\pi}{6}$ $(\because\tan(2\pi-\theta)=-\tan\theta)$ $=\frac{-1}{\sqrt{3}}$
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Find the value of the following trigonomentric ratio: $\cos\Big(-\frac{25\pi}{4}\Big)$
AnswerWe have: $\cos\Big(-\frac{25\pi}{4}\Big)=\cos(1125^\circ)$ $\cos(-1125^\circ)=\cos(1125^\circ)$ $=\cos(90^\circ\times12+45^\circ)$ 1125° lies in the first quadrant in which the cosine function is positive. Also, 12 is an integer. $\therefore\cos(-1125^\circ)=\cos(1125^\circ)$ $=\cos(90^\circ\times12+45^\circ)=\cos45^\circ=\frac{1}{\sqrt{2}}$
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Find the value of the following trigonomentric ratio: $\cos\Big(\frac{19\pi}{6}\Big)$
Answer$\cos570^\circ=\cos\Big(3\pi+\frac{\pi}{6}\Big)$ $=\cos\Big(2\pi+\Big(\pi+\frac{\pi}{6}\Big)\Big)$ $=\cos\Big(\pi+\frac{\pi}{6}\Big)$ $\begin{pmatrix}\because\cos(2\pi+\theta)=\cos\theta, \text{as cosine}\\\text{is periodoc with period }2\pi\end{pmatrix}$ $=-\cos\frac{\pi}{6}$ $(\because\cos(\pi+\theta)=-\cos\theta)$ $=\frac{-\sqrt{3}}{2}$
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Prove that: $\tan\frac{5\pi}{4}\cot\frac{9\pi}{4}+\tan\frac{17\pi}{4}\cot\frac{15\pi}{4}=0$
Answer$\text{L.H.S}=\tan\frac{5\pi}{4}\cot\frac{9\pi}{4}+\tan\frac{17\pi}{4}\cot\frac{15\pi}{4}$ $=\tan225^\circ\cot405^\circ+\tan765^\circ\cot675^\circ$ $=\tan\Big(\pi+\frac{\pi}{4}\Big)\cot\Big(2\pi+\frac{\pi}{4}\Big)+\tan\Big(4\pi+\frac{\pi}{4}\Big)\cot\Big(4\pi-\frac{\pi}{4}\Big)$ $=\tan\frac{\pi}{4}\cot\frac{\pi}{4}+\tan\frac{\pi}{4}\Big(-\cot\frac{\pi}{4}\Big)$ $=1.1+1.(-1)$ $=1-1$ $=0$ $=\text{R.H.S}$ $\text{Proved}$
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Find the value of the following trigonomentric ratios: $\sin17\pi$
Answer$360^\circ=17\pi$ $(\because\pi=180^\circ)$ $\therefore\sin360^\circ=\sin17\pi$ $=0$ $(\because \sin\text{n}\pi=\text{for all n }\in\text{ z})$
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Prove that: $\sec\Big(\frac{3\pi}{2}-\text{x}\Big)\sec\Big(\text{x}-\frac{5\pi}{2}\Big)+\tan\Big(\frac{5\pi}{2}+\text{x}\Big)\tan\Big(\text{x}-\frac{3\pi}{2}\Big)=-1$
Answer$\text{L.H.S}=\sec\Big(\frac{3\pi}{2}-\text{x}\Big)\sec\Big(\text{x}-\frac{5\pi}{2}\Big)+\tan\Big(\frac{5\pi}{2}+\text{x}\Big)\tan\Big(\text{x}-\frac{3\pi}{2}\Big)$ $=\sec\Big(\frac{3\pi}{2}-\text{x}\Big)\sec\Big(-\Big(\frac{5\pi}{2}-\text{x}\Big)\Big)+\tan\Big(\frac{5\pi}{2}+\text{x}\Big)\tan\Big(-\Big(\frac{3\pi}{2}-\text{x}\Big)\Big)$ $=-\text{cosec }\text{x}\sec\Big(\frac{5\pi}{2}-\text{x}\Big)-\cot\text{x}\times(-)\tan\Big(\frac{3\pi}{2}-\text{x}\Big)$ $\begin{bmatrix}\because\Big(\sec\Big(\frac{3\pi}{2}-\text{x}\Big)\Big)=-\text{cosec},\sec(-\text{x})=\sec\text{x},\tan\Big(\frac{5\pi}{2}+\text{x}\Big)\\=-\cot\text{x}\&\tan(-\text{x})=-\tan\text{x})\end{bmatrix}$ $=-\text{cosec}\text{x}\times\text{coses}\text{x}-\cot\text{x}\times(-1)\times\cot\text{x}$ $\begin{bmatrix}\Big(\because\sec\Big(\frac{5\pi}{2}-\text{x}\Big)\Big)=\text{cosec}\text{x}\\\&\tan\Big(\frac{3\pi}{2}-\text{x}\Big)=\cot\text{x}\\\end{bmatrix}$ $=-\text{cosec}^2\text{x}+\cot^2\text{x}$ $=-\text{cosec}^2+\text{cosec}^2-1$ $(\because\text{coses}^2\text{x}=1+\cot^2\text{x})$ $= - 1$ $\text{= R.H.S}$ $\text{Proved}$
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Prove the following identities: $\sec^4\text{x}− \sec^2\text{x}= \tan^4\text{x}+\tan^2\text{x}$
Answer$\text{L.H.S.} = \sec^4\text{x} - \sec^{2}\text{x}$ $=\sec^{2}\text{x}\big(\sec^{2}\text{x}-1\big)$ $=\big(1+\tan^{2}\text{x}\big)\tan^{2}\text{x}$ $\big[\because\sec^{2}\text{x}=1+\tan^{2}\text{x}\big]$ $=\tan^{2}\text{x}+\tan^{4}\text{x} $ $=\tan^{4}\text{x}+\tan^{2}\text{x}$ $=\text{R.H.S}$ $\text{L.H.S}=\text{R.H.S}$
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Prove that: $\frac{\tan\big(\frac{\pi}{2}-\text{x}\big)\sec(\pi-\text{x})\sin(-\text{x})}{\sin(\pi+\text{x})\cot(2\pi-\text{x})\text{cosec}\big(\frac{\pi}{2}-\text{x}\big)}=1$
Answer$\text{L.H.S}=\frac{\tan\big(\frac{\pi}{2}-\text{x}\big)\sec(\pi-\text{x})\sin(-\text{x})}{\sin(\pi+\text{x})\cot(2\pi-\text{x})\text{cosec}\big(\frac{\pi}{2}-\text{x}\big)}$ $=\frac{\cot\text{x}\times(-\sec\text{x})\times(-\sin\text{x})}{-\sin\text{x}\times(-\cot\text{x})\times\sec\text{x}}$ $= 1$ $=\text{ R.H.S}$ $\text{Proved}$
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Find the value of the following trigonomentric ratio: $\tan\Big(-\frac{13\pi}{4}\Big)$
Answer$\tan(-585^\circ)=-\tan(585)$ $(\because\tan(-\theta)=-\tan\theta)$ $=-\tan\Big(3\pi+\frac{\pi}{4}\Big)$ $=-\tan\Big(2\pi+\Big(\pi+\frac{\pi}{4}\Big)\Big)$ $(\because\tan(2\pi+\theta)=\tan\theta)$ $=-\tan\frac{\pi}{4}$ $(\because\tan({\pi+\theta})=\tan\theta)$ $=-1$
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Find the value of the following trigonomentric ratio: $\sin\Big(\frac{41\pi}{4}\Big)$
Answer$\sin1845^\circ=\sin\Big(10\pi+\frac{\pi}{4}\Big)$ $=\Big(2\times5\pi+\frac{\pi}{4}\Big)$ $=\sin\pi$ $(\because\sin ( 2\text{k}\pi+\theta)=\sin\theta,\text{ for all k}\in\text{N})$ $=\frac{1}{\sqrt{2}}$
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Find the value of the following trigonomentric ratio: $\sin\Big(-\frac{11\pi}{6}\Big)$
Answer$\sin(-330^\circ)=\sin\Big(-\Big(2\pi-\frac{\pi}{6}\Big)\Big)$ $=\sin\Big(2\pi-\frac{\pi}{6}\Big)$ $(\because\sin(-\theta=-\sin\theta))$ $=-\Big(-\sin\frac{\pi}{6}\Big)$ $(\because\sin(2\pi=\theta)=-\sin\theta)$ $=\sin\frac{\pi}{6}$ $=\frac{1}{2}$
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Prove that: $\cos24^\circ\cos55^\circ+\cos125^\circ+\cos204^\circ+\cos300^\circ=\frac12$
Answer$\text{L.H.S}=\cos24^\circ+\cos55^\circ+\cos125^\circ+\cos204^\circ+\cos300^\circ$ $=\cos24^\circ+\cos204^\circ+\cos55^\circ+\cos125^\circ+\cos300^\circ$ $=\cos24^\circ+\cos(\pi+24^\circ)+\cos55^\circ+\cos(\pi-55^\circ)+\cos\Big(2\pi-\frac{\pi}{3}\Big)$ $=\cos24^\circ-\cos24^\circ+\cos55^\circ+\cos55^\circ-\cos\frac{\pi}{3}$ $=\cos\frac{\pi}{3}$ $=\frac{1}{2}$ $=\text{R.H.S}$ $\text{Proved}$
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Find the value of the following trigonomentric ratio: $\tan\Big(\frac{7\pi}{4}\Big)$
Answer$\tan315^\circ=\tan\Big(2\pi-\frac{\pi}{4}\Big)$ $=-\tan\frac{\pi}{4}$ $(\because\tan(2\pi-\theta)=-\tan\theta)$ $=-1$
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Find the value of the following trigonomentric ratio: $\cos\Big(\frac{19\pi}{4}\Big)$
Answer$\cos(585^\circ)=\cos\Big(5\pi-\frac{\pi}{4}\Big)$ $=\cos\Big(2\times2\pi+\Big(\pi-\frac{\pi}{4}\Big)\Big)$ $=\cos\Big(\pi-\frac{\pi}{4}\Big)$ $(\because\cos(2\text{k}\pi+\theta)=\cos\theta\text{ for all k}\in\text{N})$ $=-\cos\frac{\pi}{4}$ $(\because\cos(\pi-\theta)=-\cos\theta)$ $=\frac{-1}{\sqrt{2}}$
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Prove that: $\sin\frac{13\pi}{3}\sin\frac{8\pi}{3}+\cos\frac{2\pi}{3}\sin\frac{5\pi}{6}=\frac{1}{2}$
Answer$\text{L.H.S}=\sin\frac{13\pi}{3}\sin\frac{8\pi}{3}+\cos\frac{2\pi}{3}\sin\frac{5\pi}{6}$ $=\sin780^\circ\sin480^\circ+\cos120^\circ\sin150^\circ$ $=\sin\Big(4\pi+\frac{\pi}{3}\Big)\sin\Big(3\pi+\frac{\pi}{3}\Big)+\cos\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)\sin\Big(\pi-\frac{\pi}{6}\Big)$ $(\because\pi=180^\circ)$ $=\sin\frac{\pi}{3}\times\sin\frac{\pi}{3}+\Big(-\sin\frac{\pi}{6}\Big)\sin\frac{\pi}{6}$ $\begin{pmatrix}\because\sin\Big(4\pi+\frac{\pi}{3}\Big)=\sin\frac{\pi}{3}\\\&\sin\Big(3\pi-\frac{\pi}{3}\Big)=\sin\frac{\pi}{3}\end{pmatrix}$ $=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}-\frac{1}{2}\times\frac{1}{2}$ $=\frac{3}{4}-\frac{1}{4}$ $=\frac{2}{4}$ $=\frac{1}{2}$ $=\text{R.H.S}$ $\text{Proved}$
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Prove that: $\frac{\cos(2\pi+\text{x})\text{cosec}(2\pi+\text{x})\tan\Big(\frac{\pi}{2}+\text{x}\Big)}{\sec\Big(\frac{\pi}{2}+\text{x}\Big)\cos\text{x}\cot(\pi+\text{x})}=1$
Answer$\text{L.H.S}=\frac{\cos(2\pi+\text{x})\text{cosec}(2\pi+\text{x})\tan\Big(\frac{\pi}{2}+\text{x}\Big)}{\sec\Big(\frac{\pi}{2}+\text{x}\Big)\cos\text{x}\cot(\pi+\text{x})}$ $=\frac{\cos\text{x}\times\text{cosec }\text{x}(-\cot\text{x})}{-\text{cosec }\text{x}.\cos\text{x}\cot\text{x}}$ $\begin{pmatrix}\because\tan\Big(\frac{\pi}{2}+\text{x}\Big)=-\cot\text{x}\\\&\sec\Big(\frac{\pi}{2}+\text{x}\Big)=-\text{cosec}\text{x}\end{pmatrix}$ $= 1$ $= \text{R.H.S}$ $\text{Proved}$
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In a $\triangle\text{ABC},$ Prove that: $\cos(\text{A+B})= \cos (\pi-\text{C})=0$
AnswerWe have: $\text{A + B + C}= \pi$ $\big(\because $ sum of 3 anglesof a triangle is $\pi=180^\circ\big)$ $\Rightarrow \text{A+B}=\pi-\text{C}$ $\Rightarrow \cos(\text{A+B})= \cos (\pi-\text{C})$ $\Rightarrow -\cos\text{C}$ $(\because\cos(\pi-\theta)=-\cos\theta)$ $\Rightarrow \cos(\text{A+B})+\cos\text{C}=0$ $\text{Proved}$
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Find the value of the following trigonomentric ratios: $\sin\frac{5\pi}{3}$
Answer$\sin\frac{5\pi}{3}=\sin\Big(2\pi-\frac{\pi}{3}\Big)$ $=-\sin\frac{\pi}{3}$ $(\because\sin(2\pi-\theta)=-\sin\theta)$ $=\frac{-\sqrt{3}}{2}$
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Find the value of the following trigonomentric ratio: $\sin\Big(\frac{151\pi}{6}\Big)$
Answer$4530^\circ=\Big(25\pi+\frac{\pi}{6}\Big)$ $\therefore\sin4530=\sin\Big(25\pi+\frac{\pi}{6}\Big)$ $=\sin\Big(2\times12\pi+\Big(\pi+\frac{\pi}{6}\Big)\Big)$ $=\sin\Big(\pi\frac{\pi}{6}\Big)$ $(\because\sin(2\text{k}\pi+\theta)=\sin\theta,\text{k}\in\text{N})$ $=-\sin\frac{\pi}{6}$ $(\because\sin(\pi+\theta)=-\sin\theta)$ $$$=\frac{-1}{2}$
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