9. Compound Angles Solutions

  1. Given, sinα=35\sin\alpha = \frac{3}{5} and cosβ=941\cos\beta = \frac{9}{41}

    Therefore, cosα=45\cos\alpha = \frac{4}{5} and sinβ=4041\sin\beta = \frac{40}{41}

    sin(αβ)=sinαcosβcosαsinβ\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta

    =35941454041= \frac{3}{5}\frac{9}{41} - \frac{4}{5}\frac{40}{41}

    =27160205=133205= \frac{27 - 160}{205} = -\frac{133}{205}

    cos(α+β)=cosαcosβsinαsinβ\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta

    =45941354041= \frac{4}{5}\frac{9}{41} - \frac{3}{5}\frac{40}{41}

    =36120205=84205= \frac{36 - 120}{205} = -\frac{84}{205}

  2. Given, sinα=4553\sin\alpha = \frac{45}{53} and sinβ=3365\sin\beta = \frac{33}{65}

    Thus, cosα=2853\cos\alpha = \frac{28}{53} and cosβ=5665\cos\beta = \frac{56}{65}

    sin(αβ)=sinαcosβcosαsinβ\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta

    =4553566528533365= \frac{45}{53}\frac{56}{65} - \frac{28}{53}\frac{33}{65}

    =25209243445=15963445= \frac{2520 - 924}{3445} = \frac{1596}{3445}

    sin(α+β)=sinαcosβ+cosαsinβ\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

    =45535665+28533365= \frac{45}{53}\frac{56}{65} + \frac{28}{53}\frac{33}{65}

    =2520+9243445=34443445= \frac{2520 + 924}{3445} = \frac{3444}{3445}

  3. Given, sinα=1517\sin\alpha = \frac{15}{17} and cosβ=1213\cos\beta = \frac{12}{13}

    cosα=817\cos\alpha = \frac{8}{17} and sinβ=513\sin\beta = \frac{5}{13}

    tanα=158\tan\alpha = \frac{15}{8} and tanβ=512\tan\beta = \frac{5}{12}

    sin(α+β)=sinαcosβ+cosαsinβ\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

    =15171213+817513= \frac{15}{17}\frac{12}{13} + \frac{8}{17}\frac{5}{13}

    =220221= \frac{220}{221}

    cos(αβ)=cosαcosβ+sinαsinβ\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta

    =8171213+1517513= \frac{8}{17}\frac{12}{13} + \frac{15}{17}\frac{5}{13}

    =171221= \frac{171}{221}

    tan(α+β)=tanα+tanβ1tanαtanβ\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

    =158+5121158512= \frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8}\frac{5}{12}}

    =22021= \frac{220}{21}

  4. L.H.S. =cos(45A)cos(45B)sin(45A)sin(45B)= \cos(45^{\circ} - A)\cos(45^{\circ} - B) - \sin(45^{\circ} - A)\sin(45^{\circ} - B)

    =[(cos45cosA+sin45sinA)(cos45cosB+sin45sinB)(sin45cosAcos45sinA)(sin45cosBcos45sinB)]= [(\cos 45^\circ\cos A + \sin45^\circ\sin A)(\cos 45^\circ\cos B + \sin45^\circ\sin B) - (\sin45^\circ\cos A - \cos45^\circ\sin A)(\sin45^\circ\cos B - \cos45^\circ\sin B)]

    Substituting valus for sin45\sin45^\circ and cos45\cos45^\circ

    =[(cosA2+sinA2)(cosB2+sinB2)][(cosA2sinA2)(cosB2sinB2)]=\left[\left(\frac{\cos A}{\sqrt{2}} + \frac{\sin A}{\sqrt{2}}\right)\left(\frac{\cos B}{\sqrt{2}} + \frac{\sin B}{\sqrt{2}}\right)\right] - \left[\left(\frac{\cos A}{\sqrt{2}} - \frac{\sin A}{\sqrt{2}}\right)\left(\frac{\cos B}{\sqrt{2}} - \frac{\sin B}{\sqrt{2}}\right)\right]

    =[cosAcosB2+cosAsinB2+sinAcosB2+sinAsinB2][cosAcosB2cosAsinB2sinAcosB2+sinAsinB2]= \left[\frac{\cos A\cos B}{2} + \frac{\cos A\sin B}{2} + \frac{\sin A\cos B}{2} + \frac{\sin A\sin B}{2}\right] - \left[\frac{\cos A\cos B}{2} - \frac{\cos A\sin B}{2} - \frac{\sin A\cos B}{2} + \frac{\sin A\sin B}{2}\right]

    =sinAcosB+cosAsinB=sin(A+B)= \sin A\cos B + \cos A\sin B = \sin(A + B)

  5. L.H.S. =sin(45+A)cos(45B)+cos(45+A)sin(45B)= \sin(45^{\circ} + A)\cos(45^\circ - B) + \cos(45^{\circ} + A)\sin(45^\circ - B)

    =[(sin45cosA+cos45sinA)(cos45cosB+sin45sinB)+(cos45cosAsin45sinA)(sin45cosBcos45)sinB]= [(\sin45^\circ\cos A + \cos45^\circ\sin A)(\cos45^\circ\cos B + \sin45^\circ\sin B) + (\cos45^\circ\cos A - \sin45^\circ\sin A)(\sin45^\circ\cos B - \cos45^\circ)\sin B]

    =[(cosA2+sinA2)(cosB2+sinB2)+(cosA2sinA2)(cosB2sinB2)]= \left[\left(\frac{\cos A}{\sqrt{2}} + \frac{\sin A}{\sqrt{2}}\right)\left(\frac{\cos B}{\sqrt{2}} + \frac{\sin B}{\sqrt{2}}\right) + \left(\frac{\cos A}{\sqrt{2}} - \frac{\sin A}{\sqrt{2}}\right)\left(\frac{\cos B}{\sqrt{2}} - \frac{\sin B}{\sqrt{2}}\right)\right]

    =[cosAcosB2+cosAsinB2+sinAcosB2+sinAsinB2+cosAcosB2cosAsinB2sinAcosB2+sinAsinB2]= \left[\frac{\cos A\cos B}{2} + \frac{\cos A\sin B}{2} + \frac{\sin A\cos B}{2} + \frac{\sin A\sin B}{2} + \frac{\cos A\cos B}{2} - \frac{\cos A\sin B}{2} - \frac{\sin A\cos B}{2} + \frac{\sin A\sin B}{2}\right]

    =cosAcosB+sinAsinB=cos(AB)= \cos A\cos B + \sin A\sin B = \cos(A - B)

  6. L.H.S. =sin(AB)cosAcosB+sin(BC)cosBcosC+sin(CA)cosCcosA= \frac{\sin(A - B)}{\cos A\cos B} + \frac{\sin(B - C)}{\cos B\cos C} + \frac{\sin(C - A)}{\cos C\cos A}

    =sinAcosBcosAsinBcosAcosB+sinBcosCcosBsinCcosBcosC+sinCcosAcosCsinAcosCcosA= \frac{\sin A\cos B - \cos A\sin B}{\cos A\cos B} + \frac{\sin B\cos C - \cos B\sin C}{\cos B\cos C} + \frac{\sin C\cos A - \cos C\sin A}{\cos C\cos A}

    =tanAtanB+tanBtanC+tanCtanA=0== \tan A - \tan B + \tan B - \tan C + \tan C - \tan A = 0 = R.H.S.

  7. L.H.S. =sin105+cos105= \sin 105^\circ + \cos 105^\circ

    =sin(60+45)+cos(60+45)= \sin(60^\circ + 45^\circ) + \cos(60^\circ + 45^\circ)

    =sin60cos45+cos60sin45+cos60cos45sin60sin45=\sin60^\circ\cos45^\circ + \cos60^\circ\sin45^\circ + \cos60^\circ\cos45^\circ - \sin60^\circ\sin45^\circ

    =cos45(sin60+cos60+cos60sin60)[sin45=cos45]=\cos45^\circ(\sin60^\circ + \cos60^\circ + \cos60^\circ - \sin 60^\circ)[\because \sin45^\circ = \cos45^\circ]

    =cos45== \cos45^\circ = R.H.S.

  8. Given, sin75sin15=cos105+cos15\sin 75^\circ - \sin 15^\circ = \cos 105^\circ + \cos 15^\circ

    sin75sin15=cos(90+15)+sin(9015)\Rightarrow \sin75^\circ -\sin15^\circ = \cos(90^\circ + 15^\circ) + \sin(90^\circ - 15^\circ)

    sin75sin15=cos90cos15sin90sin15+sin(9015)\Rightarrow \sin75^\circ -\sin15^\circ = \cos90^\circ\cos15^\circ - \sin90^\circ\sin15^\circ + \sin(90^\circ - 15^\circ)

    sin75sin15=sin15+sin75[cos90=0 & sin90=1]\Rightarrow \sin75^\circ -\sin15^\circ = -\sin15^\circ + \sin75^\circ [\because \cos90^\circ = 0~\&~\sin90^\circ = 1]

    Thus, we have proven the equality.

  9. L.H.S. =cosαcos(γα)sinαsin(γα)= \cos\alpha\cos(\gamma - \alpha) - \sin\alpha\sin(\gamma - \alpha)

    =cosα(cosγcosα+sinγsinα)sinα(sinγcosαsinαcosγ)= \cos\alpha(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) - \sin\alpha(\sin\gamma\cos\alpha - \sin\alpha\cos\gamma)

    =cos2αcosγ+sinγsinαcosαsinαsinγcosαsin2αcosγ= \cos^2\alpha\cos\gamma + \sin\gamma\sin\alpha\cos\alpha - \sin\alpha\sin\gamma\cos\alpha - \sin^2\alpha\cos\gamma

    =cosγ(sin2α+cos2α)=cosγ== \cos\gamma(\sin^2\alpha + \cos^2\alpha) = \cos\gamma = R.H.S.

  10. L.H.S. =cos(α+β)cosγcos(β+γ)cosα= \cos(\alpha + \beta)\cos\gamma - \cos(\beta + \gamma)\cos\alpha

    =cosαcosβcosγsinαsinβcosγcosαcosβcosγ+cosαsinβsinγ= \cos\alpha\cos\beta\cos\gamma - \sin\alpha\sin\beta\cos\gamma - \cos\alpha\cos\beta\cos\gamma + \cos\alpha\sin\beta\sin\gamma

    =sinβ(cosαsinγsinαcosγ)= \sin\beta(\cos\alpha\sin\gamma - \sin\alpha\cos\gamma)

    =sinβsin(γα)== \sin\beta\sin(\gamma - \alpha) = R.H.S.

  11. L.H.S. =sin(n+1)Asin(n1)A+cos(n+1)Acos(n1)A= \sin(n + 1)A\sin(n - 1)A + \cos(n + 1)A\cos(n - 1)A

    =cos(n+1(n1))A=cos2A== \cos(n + 1 - (n - 1))A = \cos 2A = R.H.S.

  12. L.H.S. =sin(n+1)Asin(n+2)A+cos(n+1)Acos(n+2)A= \sin(n + 1)A\sin(n + 2)A + \cos(n + 1)A\cos(n + 2)A

    =cos(n+1(n+1))=cosA== \cos(n + 1 - (n + 1)) = \cos A = R.H.S.

  13. cos15=cos(45cos30)\cos 15^\circ = \cos(45^\circ - \cos 30^\circ)

    =cos45cos30+sin45sin30= \cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ

    =1232+1212= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2}

    =3+122= \frac{\sqrt{3} + 1}{2\sqrt{2}}

    sin105=sin(60+45)\sin 105^\circ = \sin(60^\circ + 45^\circ)

    =sin60cos45+cos60sin45= \sin60^\circ\cos^45\circ + \cos60^\circ\sin45^\circ

    =3212+1212= \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}} + \frac{1}{2}\frac{1}{\sqrt{2}}

    =3+122== \frac{\sqrt{3} + 1}{2\sqrt{2}} = R.H.S.

  14. tan105=tan(60+45)\tan 105^\circ = \tan(60^\circ + 45^\circ)

    =tan60+tan451tan60tan45= \frac{\tan60^\circ + \tan 45^\circ}{1 - \tan60^\circ\tan 45^\circ}

    =3+113= \frac{\sqrt{3} + 1}{1 - \sqrt{3}}

  15. tan495cot855=tan(360+135)cot(720+135)\frac{\tan 495^\circ}{\cot 855^\circ} = \frac{\tan(360^\circ + 135^\circ)}{\cot(720^\circ + 135^\circ)}

    =tan135cot135=tan2135=(1)2=1= \frac{\tan 135^\circ}{\cot 135^\circ} = \tan^2135^\circ = (-1)^2 = 1

  16. sin(π+θ)=sinθsin(nπ+θ)=(1)nsinθ\sin(\pi + \theta) = -\sin\theta \therefore \sin(n\pi + \theta) = (1)^n\sin\theta

    sin(nπ+(1)nπ4)=(1)nsin((1)nπ4)\sin\left(n\pi + (-1)^n \frac{\pi}{4}\right) = (-1)^n\sin\left((-1)^n\frac{\pi}{4}\right)

    =(1)n(1)nsinπ4 [sin(θ)=sinθ]= (-1)^n(-1)^n\sin \frac{\pi}{4}~[\because \sin(-\theta) = -\sin\theta]

    =(1)2nsinπ4=12= (-1)^{2n}\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}

  17. L.H.S. =sin15=sin(6045)= \sin 15^\circ = \sin(60^\circ - 45^\circ)

    =sin60cos45cos60sin45= \sin60^\circ\cos45^\circ - \cos60^\circ\sin45^\circ

    =32121212= \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}} - \frac{1}{2}\frac{1}{\sqrt{2}}

    =3122== \frac{\sqrt{3} - 1}{2\sqrt{2}} = R.H.S.

  18. L.H.S. =cos75=cos(45+30)= \cos 75^\circ = \cos(45^\circ + 30^\circ)

    =cos45cos30sin45sin30= \cos45^\circ\cos30^\circ - \sin45^\circ\sin30^\circ

    =12321212= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\frac{1}{2}

    =3122== \frac{\sqrt{3} - 1}{2\sqrt{2}} = R.H.S.

  19. L.H.S. =tan75=tan(45+30)= \tan 75^\circ = \tan(45^\circ + 30^\circ)

    =tan45+tan301tan45tan30= \frac{\tan45^\circ + \tan30^\circ}{1 - \tan45^\circ\tan30^\circ}

    =1+1311.13= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1.\frac{1}{\sqrt{3}}}

    =3+13313= \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}

    =3+131=(3+1)231=2+3== \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = 2 + \sqrt{3} = R.H.S.

  20. L.H.S. =tan15=tan(4530)= \tan 15^\circ = \tan(45^\circ - 30^\circ)

    =tan45tan301+tan45tan30= \frac{\tan45^\circ - \tan30^\circ}{1 + \tan45^\circ\tan30^\circ}

    =1131+1.13= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1.\frac{1}{\sqrt{3}}}

    =3133+13= \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}

    =313+1=(31)231=23== \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = 2 - \sqrt{3} = R.H.S.

  21. cos1395=cos(3360+315)=cos315=cos(270+45)\cos 1395^\circ = \cos(3*360^\circ + 315^\circ) = \cos315^\circ = \cos(270^\circ + 45^\circ)

    =cos45=12= \cos45^\circ = \frac{1}{\sqrt{2}}

  22. tan(330)=tan(330)=tan(270+60)==cot60=13\tan(-330^\circ) = -\tan(330^\circ) = -\tan(270^\circ + 60^\circ) = =\cot 60^\circ = \frac{1}{\sqrt{3}}

  23. Given, sin300cosec1050tan(120)\sin 300^\circ \cosec 1050^\circ - \tan(-120^\circ)

    =sin(270+30)cosec(720+270+60)+tan(90+30)= \sin (270^\circ + 30^\circ)\cosec(720^\circ + 270^\circ + 60^\circ) + \tan(90^\circ + 30^\circ)

    =cos30.1cos60cot30= -cos 30^\circ. \frac{1}{-cos 60^\circ} - \cot 30^\circ

    =32.213=0= \frac{\sqrt{3}}{2}.\frac{2}{1} - \sqrt{3} = 0

  24. Given, tan(11π12)\tan\left(\frac{11\pi}{12}\right)

    =tan(ππ12)=tan15= \tan\left(\pi - \frac{\pi}{12}\right) = -\tan15^\circ

    Using the value computed in 20 for tan15\tan15^\circ we have 32\sqrt{3} - 2 as the answer.

  25. We know that tan(θ)=tanθ,tan(-\theta) = -\tan\theta, thus

    tan((1)nπ4)=(1)ntanπ4=(1)n\tan \left((-1)^n\frac{\pi}{4}\right) = (-1)^n\tan\frac{\pi}{4} = (-1)^n

  26. Given, cos18sin18=2sin27\cos 18^\circ - \sin 18^\circ = \sqrt{2}\sin 27^\circ

    12cos1812sin18=sin27\frac{1}{\sqrt{2}}\cos 18^\circ - \frac{1}{\sqrt{2}}\sin18^\circ = \sin 27^\circ

    L.H.S. =sin45cos18cos45sin18= \sin45^\circ\cos18^\circ - \cos45^\circ\sin18^\circ

    =sin(4518)=sin27== \sin(45^\circ - 18^\circ) = \sin 27^\circ = R.H.S.

  27. L.H.S. =tan70=tan(50+20)=\tan 70^\circ = \tan(50^\circ + 20^\circ)

    =tan50+tan201tan50tan20= \frac{\tan 50^\circ + \tan 20^\circ}{1 - \tan50^\circ\tan20^\circ}

    tan70tan70tan50tan20=tan50+tan20\tan70^\circ - \tan70^\circ\tan50^\circ\tan20^\circ = \tan 50^\circ + \tan 20^\circ

    tan70=tan70tan50tan20+tan50+tan20\tan70^\circ = \tan70^\circ\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ

    =tan(9020)tan50tan20+tan50+tan20= \tan(90^\circ - 20^\circ)\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ

    =cot20tan50tan20+tan50+tan20= \cot20^\circ\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ

    =tan50+tan50+tan20=2tan50+tan20== \tan 50^\circ + \tan 50^\circ + \tan 20^\circ = 2\tan50^\circ + \tan20^\circ = R.H.S.

  28. L.H.S. =cos(π4+x)cos(π4x)sin(π4+x)sin(π4x)= \frac{\cos\left(\frac{\pi}{4} + x\right)\cos\left(\frac{\pi}{4} - x\right)}{\sin\left(\frac{\pi}{4} + x\right)\sin\left(\frac{\pi}{4} - x\right)}

    =cos2π4sin2xsin2π4sin2x=12sin2x12sin2x=1== \frac{\cos^2\frac{\pi}{4} - \sin^2x}{\sin^2\frac{\pi}{4} - \sin^2x} = \frac{\frac{1}{2} - \sin^2x}{\frac{1}{2} - \sin^2x} = 1 = R.H.S.

  29. L.H.S. =cos(m+n)θ.cos(mn)θsin(m+n)θsin(mn)θ= \cos(m + n)\theta.\cos(m - n)\theta - \sin(m + n)\theta\sin(m - n)\theta

    =cos(m+n+mn)θ=cos2mθ== \cos(m + n + m - n)\theta = \cos2m\theta = R.H.S.

  30. L.H.S. =tan(θ+ϕ)+tan(θϕ)1tan(θ+ϕ)tan(θϕ)= \frac{\tan(\theta + \phi) + \tan(\theta - \phi)}{1 - \tan(\theta + \phi)\tan(\theta - \phi)}

    =tan(θ+ϕ+θϕ)=tan2θ== \tan(\theta + \phi + \theta - \phi) = \tan 2\theta = R.H.S.

  31. Given cos9+sin9=2sin54\cos 9^\circ + \sin 9^\circ = \sqrt{2}\sin 54^\circ

    12cos9+12sin9=sin54\frac{1}{\sqrt{2}}\cos9^\circ + \frac{1}{\sqrt{2}}\sin9^\circ = \sin54^\circ

    L.H.S. =sin45cos9+cos45sin9= \sin45^\circ\cos9^\circ + \cos45^\circ\sin9^\circ

    =sin(45+9)=sin54== \sin(45^\circ + 9^\circ) = \sin 54^\circ = R.H.S.

  32. L.H.S. =cos20sin20cos20+sin20= \frac{\cos 20^\circ - \sin 20^\circ}{\cos 20^\circ + \sin 20^\circ}

    Dividing both numerator and denominaor with cos20,\cos20^\circ, we get

    =1tan201+tan20=tan45tan201tan45tan20 [tan45=1]= \frac{1 - \tan20^\circ}{1 + \tan20^\circ} = \frac{\tan 45^\circ - \tan20^\circ}{1 - \tan45^\circ\tan20^\circ}~[\because \tan45^\circ = 1]

    =tan(4520)=tan25== \tan(45^\circ - 20^\circ) = \tan 25^\circ = R.H.S.

  33. L.H.S. =tanA+tanBtanAtanB= \frac{\tan A + \tan B}{\tan A - \tan B}

    =sinAcosA+sinBcosBsinAcosAsinBcosB= \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}

    =sinAcosB+sinBcosAsinAcosBsinBcosA=sin(A+B)sin(AB)== \frac{\sin A\cos B + \sin B\cos A}{\sin A\cos B - \sin B\cos A} = \frac{\sin(A + B)}{\sin (A - B)} = R.H.S.

  34. L.H.S. =1tan3AtanA1cot3AcotA= \frac{1}{\tan 3A - \tan A} - \frac{1}{\cot 3A - \cot A}

    =1tan3AtanA11tan3A1tanA= \frac{1}{\tan 3A - \tan A} - \frac{1}{\frac{1}{\tan 3A} - \frac{1}{\tan A}}

    =1tan3AtanAtanAtan3AtanAtan3A= \frac{1}{\tan 3A - \tan A} - \frac{\tan A\tan 3A}{\tan A - \tan 3A}

    =1+tanAtan3Atan3AtanA=1tan(3AA)=cot2A== \frac{1 + \tan A \tan 3A}{\tan 3A - \tan A} = \frac{1}{\tan(3A - A)} = \cot 2A = R.H.S.

  35. This is similar to previous problema and can be solved likewise.

  36. L.H.S. =sin3αsinα+cos3αcosα= \frac{\sin 3\alpha}{\sin\alpha} + \frac{\cos 3\alpha}{cos\alpha}

    =sin3αcosα+cos3αsinαsinαcosα= \frac{\sin3\alpha\cos\alpha + \cos3\alpha\sin\alpha}{\sin\alpha\cos\alpha}

    =sin(3α+α)sinαcosα= \frac{\sin(3\alpha + \alpha)}{\sin\alpha\cos\alpha}

    =22sin4αsin2α [sinαcosα=12sin2α]=2\frac{2\sin 4\alpha}{\sin2\alpha}~[\because \sin\alpha\cos\alpha = \frac{1}{2}\sin2\alpha]

    =22sin2αcos2αsin2α=4cos2α== 2\frac{2\sin2\alpha\cos2\alpha}{\sin2\alpha} = 4\cos2\alpha = R.H.S.

  37. L.H.S. =tan(π4+A)tan(π4A)tan(π4+A)+tan(π4A)= \frac{\tan\left(\frac{\pi}{4} + A \right) - \tan\left(\frac{\pi}{4} - A\right)}{\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right)}

    =1+tanA1tanA1tanA1+tanA1+tanA1tanA+1tanA1+tanA= \frac{\frac{1 + \tan A}{1 - \tan A} - \frac{1 - \tan A}{1 + \tan A}}{\frac{1 + \tan A}{1 - \tan A} + \frac{1 - \tan A}{1 + \tan A}}

    =(1+tanA)2(1tanA)2(1+tanA)2+(1tanA)2= \frac{(1 + \tan A)^2 - (1 - \tan A)^2}{(1 + \tan A)^2 + (1 - \tan A)^2}

    =4tanA2+2tan2A=2tanAsec2A=2sinAcosA=sin2A= \frac{4\tan A}{2 + 2\tan^2A} = \frac{2\tan A}{\sec^2A} = 2\sin A\cos A = \sin 2A

  38. Given, tan40+2tan10=tan50\tan 40^\circ + 2 \tan 10^\circ = \tan 50^\circ

    R.H.S. =tan50=tan(40+10)= \tan 50^\circ = \tan(40^\circ + 10^\circ)

    =tan40+tan101tan40tan10= \frac{\tan 40^\circ + \tan 10^\circ}{1 - \tan 40^\circ\tan 10^\circ}

    tan50tan50tan40tan10=tan40+tan10\tan50^\circ - \tan50^\circ\tan40^\circ\tan10^\circ = \tan 40^\circ + \tan 10^\circ

    tan50cot40tan40tan10=tan40+tan10\tan50^\circ - \cot40^\circ\tan40^\circ\tan10^\circ = tan40^\circ + \tan10^\circ

    tan50=tan40+2tan10\tan50^\circ = \tan 40^\circ + 2 \tan 10^\circ

  39. R.H.S. =tan(α+β)tan(αβ)= \tan(\alpha + \beta)\tan(\alpha - \beta)

    =sin(α+β)cos(α+β)sin(αβ)cos(αβ)= \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}\frac{\sin(\alpha -\beta)}{\cos(\alpha - \beta)}

    =sinαcosβ+cosαsinβcosαcosβsinαsinβsinαcosβcosαsinβcosαcosβ+sinαsinβ= \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}\frac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\cos\alpha\cos\beta + \sin\alpha\sin\beta}

    =sin2αcos2βsin2βcos2αcos2αcos2βsin2αsin2β= \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\cos^2\alpha\cos^2\beta - \sin^2\alpha\sin^2\beta}

    =sin2α(1sin2β)sin2β(1sin2α)cos2α(1sin2β)sin2β(1cos2α)= \frac{\sin^2\alpha(1 - \sin^2\beta) - \sin^2\beta(1 - \sin^2\alpha)}{\cos^2\alpha(1 - \sin^2\beta) - \sin^2\beta(1 - \cos^2\alpha)}

    =sin2αsin2βcos2αsin2β== \frac{\sin^2\alpha - \sin^2\beta}{\cos^2\alpha - \sin^2\beta} = R.H.S.

  40. L.H.S. =tan2αtan2β=sin2αcos2αsin2βcos2β= \tan^2\alpha -\tan^2\beta = \frac{\sin^2\alpha}{\cos^2\alpha} - \frac{\sin^2\beta}{\cos^2\beta}

    =sin2αcos2βsin2βcos2αcos2αcos2β= \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\cos^2\alpha\cos^2\beta}

    =(sinαcosβ+sinβsinα)(sinαcosβsinβsinα)cos2αcos2β= \frac{(\sin\alpha\cos\beta + \sin\beta\sin\alpha)(\sin\alpha\cos\beta - \sin\beta\sin\alpha)}{\cos^2\alpha\cos^2\beta}

    =sin(α+β)sin(αβ)cos2αcos2β== \frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\cos^2\alpha\cos^2\beta} = R.H.S.

  41. L.H.S. =tan[(2n+1)π+θ]+tan[(2n+1)πθ]= \tan[(2n + 1)\pi + \theta] + \tan[(2n + 1)\pi - \theta]

    =tan(π+θ)+tan(πθ) [tan2nπ=0]= \tan(\pi + \theta) + \tan(\pi - \theta)~[\because \tan 2n\pi = 0]

    =tanθtanθ=0== \tan\theta - \tan\theta = 0 = R.H.S.

  42. L.H.S. =tan(π4+θ)tan(3π4+θ)+1= \tan\left(\frac{\pi}{4} + \theta\right)\tan\left(\frac{3\pi}{4} + \theta\right) + 1

    =tan(π4+θ)tan[π(π4θ)]+1= \tan\left(\frac{\pi}{4} + \theta\right)\tan\left[\pi - \left(\frac{\pi}{4} - \theta\right)\right] + 1

    =tan(π4+θ)tan(π4θ)+1= -\tan\left(\frac{\pi}{4} + \theta\right)\tan\left(\frac{\pi}{4} - \theta\right) + 1

    =tan(π4+θ)tan[(π2pi4θ)]+1= -\tan\left(\frac{\pi}{4} + \theta\right)\tan\left[\left(\frac{\pi}{2} - \frac{pi}{4} - \theta\right)\right] + 1

    =tan(π4+θ)cot(π4+θ)+1=1+1=0== -\tan\left(\frac{\pi}{4} + \theta\right)\cot\left(\frac{\pi}{4} + \theta\right) + 1 = -1 + 1 = 0 = R.H.S.

  43. R.H.S. =1pq(1+p2)(1+q2)= \frac{1 - pq}{\sqrt{(1 + p^2)(1 + q^2)}}

    Substituting for pp and q,q, we get

    =1tanαtanβ(1+tan2α)(1+tan2β)= \frac{1 - \tan\alpha\tan\beta}{\sqrt{(1 + \tan^2\alpha)(1 + \tan^2\beta)}}

    =cosαcosβsinαsinβcosαcosβsecαsecβ= \frac{\frac{\cos\alpha\cos\beta - \sin\alpha\sin\beta}{\cos\alpha\cos\beta}}{\sqrt{\sec\alpha\sec\beta}}

    =cos(α+β)== \cos(\alpha + \beta) = R.H.S.

  44. Given, tanβ=2sinαsinγsin(α+γ)\tan \beta = \frac{2\sin\alpha\sin\gamma}{\sin(\alpha + \gamma)}

    Inverting, we get

    2cotβ=sin(α+γ)sinαsinγ=sinαcosγ+sinγcosαsinαsinγ2\cot\beta = \frac{\sin(\alpha + \gamma)}{\sin\alpha\sin\gamma} = \frac{\sin\alpha\cos\gamma + \sin\gamma\cos\alpha}{\sin\alpha\sin\gamma}

    =cotα+cotγ= \cot \alpha + \cot \gamma

    Thus, cotα,cotβ,cotγ\cot\alpha, \cot\beta, \cot\gamma are in A.P.

  45. tan(θ+α(θα))=tan2α=tan(θ+α)tan(θα)1+tan(θ+α)tan(θα)\tan(\theta + \alpha - (\theta - \alpha)) = \tan2\alpha = \frac{\tan(\theta + \alpha) - \tan(\theta - \alpha)}{1 + \tan(\theta + \alpha)\tan(\theta - \alpha)}

    =ba1+ab= \frac{b - a}{1 + ab}

  46. tanγ=tan(α+β)=tanα+tanβ1tanαtanβ\tan\gamma = \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

    ab=cotα+cotβtanαtanβa - b = \cot\alpha + \cot\beta - \tan\alpha - \tan\beta

    =sin(α+β)sinαsinβsin(α+β)cosαβ= \frac{\sin(\alpha + \beta)}{\sin\alpha\sin\beta} - \frac{\sin(\alpha + \beta)}{\cos\alpha\beta}

    =sin(α+β)(cosαcosβsinαsinβsinαsinβcosαcosβ)= \sin(\alpha + \beta)\left(\frac{\cos\alpha\cos\beta - \sin\alpha\sin\beta}{\sin\alpha\sin\beta\cos\alpha\cos\beta}\right)

    =sin(α+β)cos(α+β)sinαsinβcosαcosβ= \frac{\sin(\alpha + \beta)\cos(\alpha + \beta)}{\sin\alpha\sin\beta\cos\alpha\cos\beta}

    ab=(tanα+tanβ)(cotα+cotβ)ab = (\tan\alpha + \tan\beta)(\cot\alpha + \cot\beta)

    =sin2(α+β)sinαsinβcosαcosβ= \frac{\sin^2(\alpha + \beta)}{\sin\alpha\sin\beta\cos\alpha\cos\beta}

    abab=tan(α+β)=tanγ\frac{ab}{a - b} = \tan(\alpha + \beta) = \tan\gamma

  47. Given, A+B=45tan(A+B)=1A + B = 45^\circ \therefore \tan(A + B) = 1

    tanA+tanB1tanAtanB=1\frac{\tan A + \tan B}{1 - \tan A\tan B} = 1

    1+tanA+tanB+tanAtanB=21 + \tan A + \tan B + \tan A \tan B = 2

    (1+tanA)(1+tanB)=2(1 + \tan A)(1 + \tan B) = 2

  48. Given, sinαsinβcosαcosβ+1=0\sin\alpha\sin\beta - \cos\alpha\cos\beta + 1 = 0

    cosαcosβsinαsinβ=1\Rightarrow \cos\alpha\cos\beta - \sin\alpha\sin\beta = 1

    cos(α+β)=1\Rightarrow \cos(\alpha + \beta) = 1

    sin(α+β)=0\Rightarrow \sin(\alpha + \beta) = 0

    1+cotαtanβ=1+cosαsinβsinαcosβ1 + \cot\alpha\tan\beta = 1 + \frac{\cos\alpha\sin\beta}{\sin\alpha\cos\beta}

    =sinαcosβ+cosαsinβsinαcosβ= \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\sin\alpha\cos\beta}

    =sin(α+β)sinαcosβ=0sinαcosβ=0= \frac{\sin(\alpha + \beta)}{\sin\alpha\cos\beta} = \frac{0}{\sin\alpha\cos\beta} = 0

  49. tanβ=nsinαcosα1nsin2α=nsinαcosαcos2α1cos2αnsin2αcos2α\tan\beta = \frac{n\sin\alpha\cos\alpha}{1 - n\sin^2\alpha} = \frac{\frac{n\sin\alpha\cos\alpha}{\cos^2\alpha}}{\frac{1}{\cos^2\alpha} - n\frac{\sin^2\alpha}{\cos^2\alpha}}

    =ntanαsec2αntan2α=ntanα1+(1n)tan2α= \frac{n\tan\alpha}{\sec^2\alpha - n\tan^2\alpha} = \frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}

    Now, tan(αβ)=tanαntanα1+(1n)tan2α1tanαntanα1+(1n)tan2α\tan(\alpha - \beta) = \frac{\tan\alpha - \frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}}{1 - \tan\alpha\frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}}

    =tanα+(1n)tan3αntanα1+(1n)tan2α+ntan2α= \frac{\tan\alpha + (1 - n)\tan^3\alpha - n\tan\alpha}{1 + (1 - n)\tan^2\alpha + n\tan^2\alpha}

    =(1n)tanα+(1n)tan3α1+tan2α= \frac{(1 - n)\tan\alpha + (1 - n)tan^3\alpha}{1 + \tan^2\alpha}

    =(1n)tanα(1+tan2α)1+tan2α=(1n)tanα= \frac{(1 - n)\tan\alpha(1 + \tan^2\alpha)}{ 1 + \tan^2\alpha} = (1 - n)\tan\alpha

  50. Given, cos(βγ)+cos(γα)+cos(αβ)=32\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = -\frac{3}{2}

    3+2cos(βγ)+2cos(γα)+2cos(αβ)=03 + 2\cos(\beta - \gamma) + 2\cos(\gamma - \alpha) + 2\cos(\alpha - \beta) = 0

    3+2(cosβcosγ+sinβsinγ)+2(cosγcosα+sinγsinα)+2(cosαcosβ+sinαsinβ)=03 + 2(\cos\beta\cos\gamma + \sin\beta\sin\gamma) + 2(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 0

    (cos2α+sin2α)+(cos2β+sin2β)+(cos2γ+sin2γ)+2(cosβcosγ+sinβsinγ)+2(cosγcosα+sinγsinα)+2(cosαcosβ+sinαsinβ)=0(\cos^2\alpha + \sin^2\alpha) + (\cos^2\beta + \sin^2\beta) + (\cos^2\gamma + \sin^2\gamma) + 2(\cos\beta\cos\gamma + \sin\beta\sin\gamma) + 2(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 0

    (cosα+cosβ+cosγ)2+(sinα+sinβ+sinγ2)=0(\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma^2) = 0

    cosα+cosβ+cosγ=sinα+sinβ+sinγ=0\cos\alpha + \cos\beta + \cos\gamma = \sin\alpha + \sin\beta + \sin\gamma = 0

  51. tan(α+β)=tanα+tanβ1tanαtanβ\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

    =mm+1+12m+11mm+112m+1= \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \frac{m}{m + 1}\frac{1}{2m + 1}}

    =2m2+m+m+12m2+3m+1n=1= \frac{2m^2 + m + m + 1}{2m^2 + 3m + 1 - n} = 1

    Thus, α+beta=π4\alpha + beta = \frac{\pi}{4}

  52. Given (cotA1)(cotB1)=2(\cot A - 1)(\cot B - 1) = 2

    cotAcotB1cotAcotB=0\cot A\cot B - 1 - \cot A - \cot B = 0

    cotAcotB1=cotA+cotBcotAcotB1cotA+cotB=1\cot A\cot B - 1 = \cot A + \cot B \Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = 1

    cot(A+B)=cot45\cot(A + B) = \cot 45^\circ

    Thus, A+B=45A + B = 45^\circ

    which we have proved in reverse.

  53. Given, tanαtanβ=x\tan\alpha - \tan\beta = x and cotβcotα=y,\cot\beta - \cot\alpha = y, we have to prove that cot(αβ)=x+yxy\cot(\alpha - \beta) = \frac{x + y}{xy}

    Let cot