QuestionsQuestion 1Without using the tables, find the values of:(i)cos(−1230∘)\cos(-1230^\circ)cos(−1230∘)(ii)tan(−1035∘)\tan(-1035^\circ)tan(−1035∘)(iii)sec(1140∘)\sec(1140^\circ)sec(1140∘)(iv)cosec(−690∘)\operatorname{cosec}(-690^\circ)cosec(−690∘)(v)cot(1320∘)\cot(1320^\circ)cot(1320∘)(vi)cos(−240∘)\cos(-240^\circ)cos(−240∘)SolutionTheoryQuestion 2Express each of the following as a trigonometric function of an angle of positive degree measure of less than 45∘45^\circ45∘.(i)cos168∘\cos 168^\circcos168∘(ii)sin192∘\sin 192^\circsin192∘(iii)cos333∘\cos 333^\circcos333∘(iv)tan213∘\tan 213^\circtan213∘(v)cos(−435∘)\cos(-435^\circ)cos(−435∘)(vi)sin219∘\sin 219^\circsin219∘(vii)tan(−597∘)\tan(-597^\circ)tan(−597∘)(viii)cos(−111∘)\cos(-111^\circ)cos(−111∘)(ix)sin(−390∘)\sin(-390^\circ)sin(−390∘)SolutionTheoryQuestion 3Prove the following:(i)sin(180∘+α)sin(90∘−α)=−sinαcosα\sin(180^\circ + \alpha)\sin(90^\circ - \alpha) = -\sin\alpha\cos\alphasin(180∘+α)sin(90∘−α)=−sinαcosα(ii)sin810∘sin630∘+cos135∘sin225∘=−12\sin 810^\circ \sin 630^\circ + \cos 135^\circ \sin 225^\circ = -\dfrac{1}{2}sin810∘sin630∘+cos135∘sin225∘=−21(iii)tan150∘cot330∘−2sec135∘cosec225∘=−3\tan 150^\circ \cot 330^\circ - 2\sec 135^\circ \operatorname{cosec} 225^\circ = -3tan150∘cot330∘−2sec135∘cosec225∘=−3(iv)sin210∘+cos240∘+tan225∘+cot225∘=1\sin 210^\circ + \cos 240^\circ + \tan 225^\circ + \cot 225^\circ = 1sin210∘+cos240∘+tan225∘+cot225∘=1SolutionTheoryQuestion 4Prove that:(i)tan(180∘+α)cot(90∘−α)sin(360∘−α)cos(270∘+α)=−sec2α\dfrac{\tan(180^\circ + \alpha)\cot(90^\circ - \alpha)}{\sin(360^\circ - \alpha)\cos(270^\circ + \alpha)} = -\sec^2\alphasin(360∘−α)cos(270∘+α)tan(180∘+α)cot(90∘−α)=−sec2α(ii)sin2(π+θ)tan(3π2+θ)cot2(3π2−θ)cos2(π−θ)cosec(2π−θ)=cosθ\dfrac{\sin^2(\pi + \theta)\tan\left(\dfrac{3\pi}{2} + \theta\right)}{\cot^2\left(\dfrac{3\pi}{2} - \theta\right)\cos^2(\pi - \theta)\operatorname{cosec}(2\pi - \theta)} = \cos\thetacot2(23π−θ)cos2(π−θ)cosec(2π−θ)sin2(π+θ)tan(23π+θ)=cosθ(iii)cos(90∘+θ)sec(−θ)tan(180∘−θ)sec(360∘−θ)sin(180∘+θ)cot(90∘−θ)=−1\dfrac{\cos(90^\circ + \theta)\sec(-\theta)\tan(180^\circ - \theta)}{\sec(360^\circ - \theta)\sin(180^\circ + \theta)\cot(90^\circ - \theta)} = -1sec(360∘−θ)sin(180∘+θ)cot(90∘−θ)cos(90∘+θ)sec(−θ)tan(180∘−θ)=−1SolutionTheoryQuestion 5Show that: sec(3π2−θ)sec(5π2−θ)−tan(3π2−θ)tan(5π2+θ)=−1\sec\left(\dfrac{3\pi}{2} - \theta\right)\sec\left(\dfrac{5\pi}{2} - \theta\right) - \tan\left(\dfrac{3\pi}{2} - \theta\right)\tan\left(\dfrac{5\pi}{2} + \theta\right) = -1sec(23π−θ)sec(25π−θ)−tan(23π−θ)tan(25π+θ)=−1.SolutionTheoryQuestion 6If α,β,γ\alpha, \beta, \gammaα,β,γ are the angles of a triangle ABCABCABC, then prove that(i)sin(α+β)=sinγ\sin(\alpha + \beta) = \sin\gammasin(α+β)=sinγ(ii)sec(α+β2)=cscγ2\sec\left(\dfrac{\alpha + \beta}{2}\right) = \operatorname{csc}\dfrac{\gamma}{2}sec(2α+β)=csc2γ(iii)cosecα=1sin(β+γ)\operatorname{cosec}\alpha = \dfrac{1}{\sin(\beta + \gamma)}cosecα=sin(β+γ)1(iv)tan(α+β)+tanγ=0\tan(\alpha + \beta) + \tan\gamma = 0tan(α+β)+tanγ=0SolutionTheory