QuestionsQuestion 1Find the values of sin2α\sin 2\alphasin2α, cos2α\cos 2\alphacos2α and tan2α\tan 2\alphatan2α, when:(i)sinα=35\sin\alpha = \dfrac{3}{5}sinα=53(ii)cosα=45\cos\alpha = \dfrac{4}{5}cosα=54, where 0<α<π20 < \alpha < \dfrac{\pi}{2}0<α<2πSolutionTheoryQuestion 2Prove the following identities:(i)cotα−tanα=2cot2α\cot\alpha - \tan\alpha = 2\cot 2\alphacotα−tanα=2cot2α(ii)sin2α1+cos2α=tanα\dfrac{\sin 2\alpha}{1 + \cos 2\alpha} = \tan\alpha1+cos2αsin2α=tanα(iii)1−cosαsinα=tanα2\dfrac{1 - \cos\alpha}{\sin\alpha} = \tan\dfrac{\alpha}{2}sinα1−cosα=tan2α(iv)cosα−sinαcosα+sinα=sec2α−tan2α\dfrac{\cos\alpha - \sin\alpha}{\cos\alpha + \sin\alpha} = \sec 2\alpha - \tan 2\alphacosα+sinαcosα−sinα=sec2α−tan2α(v)1+sinα1−sinα=sinα2+cosα2sinα2−cosα2\sqrt{\dfrac{1 + \sin\alpha}{1 - \sin\alpha}} = \dfrac{\sin\dfrac{\alpha}{2} + \cos\dfrac{\alpha}{2}}{\sin\dfrac{\alpha}{2} - \cos\dfrac{\alpha}{2}}1−sinα1+sinα=sin2α−cos2αsin2α+cos2α(vi)cosecθ+2cosec2θsecθ=cotθ2\dfrac{\operatorname{cosec}\theta + 2\operatorname{cosec} 2\theta}{\sec\theta} = \cot\dfrac{\theta}{2}secθcosecθ+2cosec2θ=cot2θ(vii)1+tanαtan2α=sec2α1 + \tan\alpha \tan 2\alpha = \sec 2\alpha1+tanαtan2α=sec2α(viii)2sinθsin2θcosθ+cos3θ=tan2θtanθ\dfrac{2\sin\theta \sin 2\theta}{\cos\theta + \cos 3\theta} = \tan 2\theta \tan\thetacosθ+cos3θ2sinθsin2θ=tan2θtanθ(ix)sin3θsinθ−cos3θcosθ=2\dfrac{\sin 3\theta}{\sin\theta} - \dfrac{\cos 3\theta}{\cos\theta} = 2sinθsin3θ−cosθcos3θ=2(x)cos3θcosθ+sin3θsinθ=4cos2θ\dfrac{\cos 3\theta}{\cos\theta} + \dfrac{\sin 3\theta}{\sin\theta} = 4\cos 2\thetacosθcos3θ+sinθsin3θ=4cos2θ(xi)tanθ2+cotθ2cotθ2−tanθ2=secθ\dfrac{\tan\dfrac{\theta}{2} + \cot\dfrac{\theta}{2}}{\cot\dfrac{\theta}{2} - \tan\dfrac{\theta}{2}} = \sec\thetacot2θ−tan2θtan2θ+cot2θ=secθ(xii)sin3θcosθ+cos3θsinθ=2cot2θ\dfrac{\sin 3\theta}{\cos\theta} + \dfrac{\cos 3\theta}{\sin\theta} = 2\cot 2\thetacosθsin3θ+sinθcos3θ=2cot2θ(xiii)3+cos4θ1−cos4θ=12(tan2θ+cot2θ)\dfrac{3 + \cos 4\theta}{1 - \cos 4\theta} = \dfrac{1}{2}(\tan^2\theta + \cot^2\theta)1−cos4θ3+cos4θ=21(tan2θ+cot2θ)(xiv)1+sin2θ1−sin2θ=tan(π4+θ)\dfrac{1 + \sin 2\theta}{1 - \sin 2\theta} = \tan\left(\dfrac{\pi}{4} + \theta\right)1−sin2θ1+sin2θ=tan(4π+θ)(xv)cos2π8+cos23π8+cos25π8+cos27π8=2\cos^2\dfrac{\pi}{8} + \cos^2\dfrac{3\pi}{8} + \cos^2\dfrac{5\pi}{8} + \cos^2\dfrac{7\pi}{8} = 2cos28π+cos283π+cos285π+cos287π=2SolutionTheoryQuestion 3Show that: 2cosθ=2+2+2cos4θ2\cos\theta = \sqrt{2 + \sqrt{2 + 2\cos 4\theta}}2cosθ=2+2+2cos4θ.SolutionTheoryQuestion 4Reduce sin4θ\sin^4\thetasin4θ to an expression involving only function of multiples of θ\thetaθ, raised to the first power.SolutionTheoryQuestion 5Find the values of sinθ\sin\thetasinθ and cosθ\cos\thetacosθ without using table or calculator, when θ\thetaθ is:(i)18∘18^\circ18∘(ii)36∘36^\circ36∘(iii)54∘54^\circ54∘(iv)72∘72^\circ72∘SolutionTheory