13. Multiple and Submultiple Angles Solutions#

  1. Let us solve these one by one.

    1. Given, cosA=35\cos A = \frac{3}{5}

      sinA=1cos2A=1925=1625=45\Rightarrow \sin A = \sqrt{1 - \cos^2A} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

      sin2A=2sinAcosA=2.45.35=2425\sin 2A = 2\sin A\cos A = 2.\frac{4}{5}.\frac{3}{5} = \frac{24}{25}

    2. Given, sinA=1213\sin A = \frac{12}{13}

      cosA=1sin2A=1144169=25169=513\Rightarrow \cos A = \sqrt{1 - \sin^2A} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}

      sin2A=2sinAcosA=2.1213.513=120169\sin 2A = 2\sin A\cos A = 2.\frac{12}{13}.\frac{5}{13} = \frac{120}{169}

    3. Given, tanA=1663=perpendicularbase\tan A = \frac{16}{63} = \frac{\text{perpendicular}}{\text{base}}

      hypotenuse=p2+b2=162+632=65\text{hypotenuse} = \sqrt{p^2 + b^2} = \sqrt{16^2 + 63^2} = 65

      sinA=1665,cosA=6365\sin A = \frac{16}{65}, \cos A = \frac{63}{65}

      sin2A=2sinAcosA=2.1665.6365=20164225\sin 2A = 2\sin A\cos A = 2.\frac{16}{65}.\frac{63}{65} = \frac{2016}{4225}

  2. Let us solve these one by one.

    1. Given, cosA=1517\cos A = \frac{15}{17}

      sinA=1cos2A\Rightarrow \sin A = \sqrt{1 - \cos^2A} =1225289=64289=817= \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17}

      cos2A=cos2Asin2A=22564289=161289\cos 2A = \cos^2A - \sin^2A = \frac{225 - 64}{289} = \frac{161}{289}

    2. Given, sinA=45\sin A = \frac{4}{5}

      cosA=1sin2A\Rightarrow \cos A = \sqrt{1 - \sin^2A} =11625=35= \sqrt{1 - \frac{16}{25}} = \frac{3}{5}

      cos2A=cos2Asin2A=91625=725\cos2A = \cos^2A - \sin^2A = \frac{9 - 16}{25} = -\frac{7}{25}

    3. Give, tanA=512=perpendicularbase\tan A = \frac{5}{12} = \frac{\text{perpendicular}}{\text{base}}

      hypotenuse=p2+b2=25+144=13\text{hypotenuse} = \sqrt{p^2 + b^2} = \sqrt{25 + 144} = 13

      sinA=513,cosA=1213\sin A = \frac{5}{13}, \cos A = \frac{12}{13}

      cos2A=cos2Asin2A=119169\cos^2A = \cos^2A - \sin^2A = \frac{119}{169}

  3. Given, tanA=ba,\tan A = \frac{b}{a}, thus hypotenuse=b2+a2\text{hypotenuse} = \sqrt{b^2 + a^2}

    acos2A+bsin2A=a(cos2Asin2A)+2bsinAcosAa\cos 2A+ b\sin 2A = a(\cos^2A - \sin^2A) + 2b\sin A\cos A

    =a(a2a2+b2b2a2+b2)+2b.aba2+b2= a\left(\frac{a^2}{a^2 + b^2} - \frac{b^2}{a^2 + b^2}\right) + 2b.\frac{ab}{a^2 + b^2}

    =a(a2b2+2b2a2+b2)=a= a\left(\frac{a^2 - b^2 + 2b^2}{a^2 + b^2}\right) = a

  4. We have to prove that sin2A1+cos2A=tanA\frac{\sin 2A}{1 + \cos 2A} = \tan A

    L.H.S. =sin2A1+cos2A=2sinAcosA1+cos2Asin2A= \frac{\sin 2A}{1 + \cos 2A} = \frac{2\sin A\cos A}{1 + \cos^2A - \sin^2A}

    =2sinAcosA2cos2A[1sin2A=cos2A]= \frac{2\sin A\cos A}{2\cos^2A}[\because 1 - \sin^2A = \cos^2A]

    =tanA== \tan A = R.H.S.

  5. We have to prove that sin2A1cos2A=cotA\frac{\sin 2A}{1 - \cos 2A} = \cot A

    L.H.S. =sin2A1cos2A=2sinAcosA1(cos2Asin2A)= \frac{\sin 2A}{1 - \cos 2A} = \frac{2\sin A\cos A}{1 -(\cos^2A - \sin^2A)}

    =2sinAcosA2sin2A=cotA== \frac{2\sin A\cos A}{2\sin^2A} = \cot A = R.H.S.

  6. We have to prove that 1cos2A1+cos2A=tan2A\frac{1 - \cos 2A}{1 + \cos 2A} = \tan^2A

    L.H.S. =1(cos2Asin2A)1+cos2Asin2A= \frac{1 - (\cos^2A - \sin^2A)}{1 + \cos^2A - \sin^2A}

    =2sin2A2cos2A=tan2A== \frac{2\sin^2A}{2\cos^2A} = \tan^2A = R.H.S.

  7. We have to prove that tanA+cotA=2cosec2A\tan A + \cot A = 2\cosec 2A

    L.H.S. =sinAcosA+cosAsinA=sin2A+cos2AsinAcosA= \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2A + \cos^2A}{\sin A\cos A}

    =22sinAcosA=2sin2A=2cosec2A== \frac{2}{2\sin A\cos A} = \frac{2}{\sin 2A} = 2\cosec 2A = R.H.S.

  8. We have to prove that tanAcotA=2cot2A\tan A - \cot A = -2\cot2A

    L.H.S. =sinAcosAcosAsinA=sin2Acos2AsinAcosA= \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A} = \frac{\sin^2A - \cos^2A}{\sin A\cos A}

    =cos2Asin2A2=2cot2A== \frac{-\cos2A}{\frac{\sin2A}{2}} = -2\cot2A = R.H.S.

  9. We have to prove that cosec2A+cot2A=cotA\cosec 2A + \cot 2A = \cot A

    L.H.S. =1sin2A+cos2Asin2A=1+cos2Asin2A=2cos2A2sinAcosA= \frac{1}{\sin 2A} + \frac{\cos 2A}{\sin 2A} = \frac{1 + \cos 2A}{\sin 2A} = \frac{2\cos^2A}{2\sin A\cos A}

    =cotA== \cot A = R.H.S.

  10. We have to prove that 1cosA+cosBcos(A+B)1+cosAcosBcos(A+B)=tanA2cotB2\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)} = \tan\frac{A}{2}\cot\frac{B}{2}

    L.H.S. =1cosA+cosBcos(A+B)1+cosAcosBcos(A+B)= \frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)}

    =2sin2A2+2sinA2sin(A2+B)2cos2A22cosA2cos(A2+B)= \frac{2\sin^2\frac{A}{2} + 2\sin\frac{A}{2}\sin\left(\frac{A}{2} + B\right)}{2\cos^2\frac{A}{2} - 2\cos\frac{A}{2}\cos\left(\frac{A}{2} + B\right)}

    =sinA2(sinA2+sin(A2+B))cosA2(cosA2cos(A2+B))= \frac{\sin\frac{A}{2}\left(\sin\frac{A}{2} + \sin \left(\frac{A}{2} + B\right)\right)}{\cos\frac{A}{2}\left(\cos\frac{A}{2} - \cos\left(\frac{A}{2} + B\right)\right)}

    =tanA2(2sin(A+B2)cosB2)2sin(A+B2)sinB2= \frac{\tan\frac{A}{2}\left(2\sin\left(\frac{A + B}{2}\right)\cos\frac{B}{2}\right)}{2\sin\left(\frac{A + B}{2}\right)\sin\frac{B}{2}}

    =tanA2cotB2= \tan\frac{A}{2}\cot\frac{B}{2}

  11. We have to prove that cosA1sinA=tan(45±A2)\frac{\cos A}{1 \mp \sin A} = \tan\left(45^\circ \pm \frac{A}{2}\right)

    First considering - sign on L.H.S.,

    L.H.S. =cosA1sinA=cos2A2sin2A2(cosA2sinA2)2= \frac{\cos A}{1 - \sin A} = \frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\left(\cos\frac{A}{2} - \sin\frac{A}{2}\right)^2}

    Dividing numerator and denomiator by cos2A2\cos^2\frac{A}{2}

    =1tan2A2(1tanA2)2= \frac{1 - \tan^2\frac{A}{2}}{\left(1 - \tan\frac{A}{2}\right)^2}

    =1+tanA21tanA2= \frac{1 + \tan\frac{A}{2}}{1 - \tan\frac{A}{2}}

    =tan45+tanA21tan45tanA2=tan(45+A2)= \frac{\tan45^\circ + \tan\frac{A}{2}}{1 - \tan45^\circ\tan\frac{A}{2}} = \tan\left(45^\circ + \frac{A}{2}\right)

    Similarly by considering the ++ sign we can prove the other sign.

  12. We have to prove that sec8A1sec4A1=tan8Atan2A\frac{\sec 8A - 1}{\sec 4A - 1} = \frac{\tan 8A}{\tan 2A}

    L.H.S. =sec8A1sec4A1=1cos8A1cos4A.cos4Acos8A= \frac{\sec 8A - 1}{\sec 4A - 1} = \frac{1 - \cos 8A}{1 - \cos 4A}.\frac{\cos4A}{\cos8A}

    =2sin24A2sin22A.cos4Acos8A=(2sin4Acos4A).sin4A2sin22A.cos8A= \frac{2\sin^24A}{2\sin^22A}.\frac{\cos 4A}{\cos8A} = \frac{(2\sin4A\cos4A).\sin4A}{2\sin^22A.\cos8A}

    =sin8Acos8A.sin4A2sin22A=tan8A.2sin2Acos2A2sin22A=tan8Atan2A== \frac{\sin8A}{\cos8A}.\frac{\sin4A}{2\sin^22A} = \frac{\tan8A. 2\sin2A\cos2A}{2\sin^22A} = \frac{\tan8A}{\tan2A} = R.H.S.

  13. We have to prove that 1+tan2(45A)1tan2(45A)=cosec2A\frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)} = \cosec 2A

    L.H.S. =1+tan2(45A)1tan2(45A)= \frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)}

    =cos2(45A)+sin2(45A)cos2(45A)sin2(45A)= \frac{\cos^2(45^\circ - A) + \sin^2(45^\circ - A)}{\cos^2(45^\circ - A) - \sin^2(45^\circ - A)}

    =1cos(902A)=1sin2A=cosec2A== \frac{1}{\cos(90^\circ - 2A)} = \frac{1}{\sin2A} = \cosec2A = R.H.S.

  14. We have to prove that sinA+sinBsinAsinB=tanA+B2tanAB2\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}}

    L.H.S. =sinA+sinBsinAsinB=2sinA+B2cosAB22cosA+B2sinAB2= \frac{\sin A + \sin B}{\sin A - \sin B} = \frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}

    =tanA+B2tanAB2== \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}} = R.H.S.

  15. We have to prove that sin2Asin2BsinAcosAsinBcosB=tan(A+B)\frac{\sin^2A - \sin^2B}{\sin A\cos A - \sin B\cos B} = \tan(A + B)

    L.H.S. =2(cos2Bcos2A)sin2Asin2B=cos2Bcos2Asin2Asin2B= \frac{2(\cos^2B - \cos^2A)}{\sin2A - \sin2B} = \frac{\cos2B - \cos2A}{\sin2A - \sin2B}

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

  16. We have to prove that tan(π4+A)tan(π4A)=2tan2A\tan\left(\frac{\pi}{4} + A\right) - \tan\left(\frac{\pi}{4} - A\right) = 2\tan 2A

    L.H.S. =1+tanA1tanA1tanA1+tanA= \frac{1 + \tan A}{1 - \tan A} - \frac{1 - \tan A}{1 + \tan A}

    =(1+tanA)2(1tanA)21tan2A=4tanA1tan2A= \frac{(1 + \tan A)^2 - (1 - \tan A)^2}{1 - \tan^2A} = \frac{4\tan A}{1 - \tan^2A}

    =4sinAcosA.cos2Acos2Asin2A=2sin2Acos2A=2tan2A== \frac{4\sin A}{\cos A}. \frac{\cos^2A}{\cos^2A - \sin^2A} = \frac{2\sin2A}{\cos2A} = 2\tan2A = R.H.S.

  17. We have to prove that cosA+sinAcosAsinAcosAsinAcosA+sinA=2tan2A\frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} = 2\tan 2A

    L.H.S. =(cosA+sinA)2(cosAsinA)2cos2Asin2A= \frac{(\cos A + \sin A)^2 - (\cos A - \sin A)^2}{\cos^2A - \sin^2A}

    =4cosAsinAcos2A=2sin2Acos2A=2tan2A== \frac{4\cos A\sin A}{\cos 2A} = \frac{2\sin 2A}{\cos 2A} = 2\tan2A = R.H.S.

  18. We have to prove that cot(A+15)tan(A15)=4cos2A1+2sin2A\cot (A + 15^\circ) - \tan(A - 15^\circ) = \frac{4\cos 2A}{1 + 2\sin 2A}

    L.H.S. =1tan(A+15)tan(A15)tan(A+15)= \frac{1 - \tan(A + 15^\circ)\tan(A - 15^\circ)}{\tan(A + 15^\circ)}

    =cos(A+15)cos(A15)sin(A+15)sin(A15)cos(A+15)cos(A15).cos(A+15)sin(A+15)= \frac{\cos(A + 15^\circ)\cos(A - 15^\circ) - \sin(A + 15^\circ)\sin(A - 15^\circ)}{\cos(A + 15^\circ)\cos(A - 15^\circ)}.\frac{\cos(A + 15^\circ)}{\sin(A + 15^\circ)}

    =cos2Asin(A+15)cos(A15)=2cos2A2sin(A+15)cos(A15)= \frac{\cos 2A}{\sin(A + 15^\circ)\cos(A - 15^\circ)} = \frac{2\cos 2A}{2\sin(A + 15^\circ)\cos(A - 15^\circ)}

    =2cos2Asin2A+sin30=4cos2A1+sin2A== \frac{2\cos 2A}{\sin2A + \sin30^\circ} = \frac{4\cos 2A}{1 + \sin 2A} = R.H.S.

  19. We have to prove that sinA+sin2A1+cosA+cos2A=tanA\frac{\sin A + \sin2A}{1 + \cos A + \cos 2A} = \tan A

    L.H.S. =sinA+2sinAcosAcosA+2cos2A=sinA(1+2cosA)cosA(1+2cosA)= \frac{\sin A + 2\sin A\cos A}{\cos A + 2\cos^2A} = \frac{\sin A(1 + 2\cos A)}{\cos A(1 + 2\cos A)}

    =tanA== \tan A = R.H.S.

  20. We have to prove that 1+sinAcosA1+sinA+cosA=tanA2\frac{1 + \sin A - \cos A }{1 + \sin A + cos A} = \tan \frac{A}{2}

    L.H.S. =2sin2A2+2sinA2cosA22cos2A2+2sinA2cosA2= \frac{2\sin^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}

    =sinA2(sinA2+cosA2)cosA2(sinA2+cosA2)= \frac{\sin\frac{A}{2}(\sin\frac{A}{2} + \cos\frac{A}{2})}{\cos\frac{A}{2}(\sin\frac{A}{2} + \cos\frac{A}{2})}

    =tanA2== \tan\frac{A}{2} = R.H.S.

  21. We have to prove that sin(n+1)Asin(n1)Acos(n+1)A+2cosnA+cos(n1)A=tanA2\frac{\sin(n + 1)A - \sin(n - 1)A}{\cos(n + 1)A + 2\cos nA + \cos(n - 1)A} = \tan \frac{A}{2}

    L.H.S. =2cosnAsinA2cosnAcosA+2cosnA=sinA1+cosA= \frac{2\cos nA \sin A}{2\cos nA \cos A + 2\cos nA} = \frac{\sin A}{1 + \cos A}

    =2sinA2cosA22cos2A2=tanA2== \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2}} = \tan \frac{A}{2} = R.H.S.

  22. We have to prove that sin(n+1)A+2sinnA+sin(n1)Acos(n1)cos(n+1)A=cotA2\frac{\sin(n + 1)A + 2\sin nA + \sin(n - 1)A}{\cos(n - 1) - \cos(n + 1)A} = \cot \frac{A}{2}

    L.H.S. =2sinnAcosA+2sinnA2sinnAsinA= \frac{2\sin nA\cos A + 2\sin nA}{2\sin nA\sin A}

    =cosA+1sinA=2cos2A22sinA2cosA2= \frac{\cos A + 1}{\sin A} = \frac{2\cos^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}

    =cotA2== \cot\frac{A}{2} = R.H.S.

  23. We have to prove that sin(2n+1)AsinA=sin2(n+1)Asin2nA\sin(2n + 1)A\sin A = \sin^2(n + 1)A - \sin^2nA

    R.H.S. =(sin(n+1)A+sinnA)(sin(n+1)AsinnA)= (\sin(n + 1)A + \sin nA)(\sin(n + 1)A - \sin nA)

    =(2sin2n+12AcosA2)(2cos2n+12AsinA2)= (2\sin\frac{2n + 1}{2}A\cos \frac{A}{2})(2\cos\frac{2n + 1}{2}A\sin \frac{A}{2})

    =2sin2n+12Acos2n+12A.2cosA2sinA2= 2\sin\frac{2n + 1}{2}A\cos\frac{2n + 1}{2}A.2\cos \frac{A}{2}\sin\frac{A}{2}

    =sin(2n+1)AsinA== \sin(2n + 1)A\sin A = L.H.S.

  24. We have to prove that sin(A+3B)+sin(3A+B)sin2A+sin2B=2cos(A+B)\frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B} = 2\cos(A + B)

    L.H.S. =sin(A+3B)+sin(3A+B)sin2A+sin2B= \frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B}

    =2sin(2A+2B)cos(AB)2sin(A+B)cos(AB)= \frac{2\sin(2A + 2B)\cos(A - B)}{2\sin(A + B)\cos(A - B)}

    =2sin(A+B)cos(A+B)sin(A+B)=2cos(A+B)== \frac{2\sin(A + B)\cos(A + B)}{\sin(A + B)} = 2\cos(A + B) = R.H.S.

  25. We have to prove that sin3A+sin2AsinA=4sinAcosA2cos3A2\sin 3A + \sin 2A - \sin A = 4\sin A\cos \frac{A}{2}\cos \frac{3A}{2}

    L.H.S. =2cos2AsinA+2sinAcosA=2sinA(cos2A+cosA)= 2\cos 2A\sin A + 2\sin A\cos A = 2\sin A(\cos 2A + \cos A)

    =2sinAcos3A2cosA2== 2\sin A\cos \frac{3A}{2}\cos\frac{A}{2} = R.H.S.

  26. We have to prove that tan2A=(sec2A+1)sec2A1\tan 2A = (\sec 2A + 1)\sqrt{\sec^2A - 1}

    R.H.S. =1+cos2Acos2A1cos2Acos2A= \frac{1 + \cos 2A}{\cos 2A}\sqrt{\frac{1 - \cos^2A}{\cos^2A}}

    =2cos2A2cos2A1.sin2Acos2A= \frac{2\cos^2A}{2\cos^2A - 1}.\sqrt{\frac{\sin^2A}{\cos^2A}}

    =22sec2A.tanA=2tanA1tan2A=tanA+tanA1tanA.tanA= \frac{2}{2 - \sec^2A}.tan A = \frac{2\tan A}{1 - \tan^2A} = \frac{\tan A + \tan A}{1 - \tan A.\tan A}

    =tan2A==\tan 2A = R.H.S.

  27. We have to prove that cos32A+3cos2A=4(cos6Asin6A)\cos^32A + 3\cos 2A = 4(\cos^6A - \sin^6A)

    L.H.S. =(cos2Asin2A)3+3(cos2Asin2A)= (\cos^2A - \sin^2A)^3 + 3(\cos^2A - \sin^2A)

    =cos6A3cos4Asin2A+3cos2Asin4Asin6A+3(cos2Asin2A)= \cos^6A -3\cos^4A\sin^2A + 3\cos^2A\sin^4A - \sin^6A + 3(\cos^2A - \sin^2A)

    =cos6A3cos4A(1cos2A)+3(1sin2A)sin4Asin6A+3(cos2Asin2A)= \cos^6A -3\cos^4A(1 - \cos^2A) + 3(1 - \sin^2A)\sin^4A - \sin^6A + 3(\cos^2A - \sin^2A)

    =4(cos6Asin6A)== 4(\cos^6A - \sin^6A) = R.H.S.

  28. We have to prove that 1+cos22A=2(cos4A+sin4A)1 + \cos^22A = 2(\cos^4A + \sin^4A)

    L.H.S. =1+(cos2Asin2A)2=12sin2Acos2A+cos4A+sin4A= 1 + (\cos^2A - \sin^2A)^2 = 1 - 2\sin^2A\cos^2A + \cos^4A + \sin^4A

    =12sin2A(1sin2A)+cos4A+sin4A= 1 - 2\sin^2A(1 - \sin^2A) + \cos^4A + \sin^4A

    =12sin2A+2sin4A+cos4A+sin4A= 1 - 2\sin^2A + 2\sin^4A + \cos^4A + \sin^4A

    =(1sin2A)2+cos4A+2sin4A=2(cos4A+sin4A)== (1 - \sin^2A)^2 + \cos^4A + 2\sin^4A = 2(\cos^4A + \sin^4A) = R.H.S.

  29. We have to prove that sec2A(1+sec2A)=2sec2A\sec^2A(1 + \sec2A) = 2\sec2A

    L.H.S. =1cos2A.cos2A+1cos2A= \frac{1}{\cos^2A}.\frac{\cos2A + 1}{\cos 2A}

    =1cos2A.2cos2Acos2A=2sec2A== \frac{1}{\cos^2A}.\frac{2\cos^2A}{\cos 2A} = 2\sec2A = R.H.S.

  30. We have to prove that cosecA2cot2AcosA=2sinA\cosec A - 2\cot 2A\cos A = 2\sin A

    L.H.S. =1sinA2cos2AcosAsin2A= \frac{1}{\sin A} - \frac{2\cos 2A\cos A}{\sin 2A}

    =1sinA2cos2AcosA2sinAcosA= \frac{1}{\sin A} - \frac{2\cos 2A\cos A}{2\sin A\cos A}

    1sinAcos2AsinA=1cos2AsinA\frac{1}{\sin A} - \frac{\cos 2A}{\sin A} = \frac{1 - \cos 2A}{\sin A}

    =2sin2AsinA=2sinA== \frac{2\sin^2A}{\sin A} = 2\sin A = R.H.S.

  31. We have to prove that cotA=12(cotA2tanA2)\cot A = \frac{1}{2}\left(\cot\frac{A}{2} - \tan\frac{A}{2}\right)

    R.H.S. =12(1tan2A2tanA2)= \frac{1}{2}\left(\frac{1 - \tan^2\frac{A}{2}}{\tan\frac{A}{2}}\right)

    =12(cos2A2sin2A2cos2A2).cosA2sinA2= \frac{1}{2}\left(\frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\cos^2\frac{A}{2}}\right).\frac{\cos\frac{A}{2}}{\sin \frac{A}{2}}

    =12cosAcosA2.1sinA2=cotA== \frac{1}{2}\frac{\cos A}{\cos\frac{A}{2}}.\frac{1}{\sin\frac{A}{2}} = \cot A = L.H.S.

  32. We have to prove that sinAsin(60A)sin(60+A)=14sin3A\sin A\sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4}\sin 3A

    L.H.S. =sinA.cos2Acos1202=sinA(12sin2A+12)2=\sin A.\frac{\cos 2A - \cos 120^\circ}{2} = \frac{\sin A\left(1 - 2\sin^2A + \frac{1}{2}\right)}{2}

    =3sinA4sin3A4=14sin3A== \frac{3\sin A - 4\sin^3A}{4} = \frac{1}{4}\sin 3A = R.H.S.

  33. We have to prove that cosAcos(60A)cos(60+A)=14cos3A\cos A\cos(60^\circ - A)\cos(60^\circ + A) = \frac{1}{4}\cos 3A

    L.H.S. =cosA2(cos2A+cos120)=cosA2(2cos2A112)= \frac{\cos A}{2}\left(\cos 2A + \cos120^\circ\right) = \frac{\cos A}{2}\left(2\cos^2A - 1 - \frac{1}{2}\right)

    =4cos3A3cosA4=14cos3A== \frac{4\cos^3A - 3\cos A}{4} = \frac{1}{4}\cos 3A = R.H.S.

  34. We have to prove that cotA+cot(60+A)cot(60A)=3cot3A\cot A + \cot(60^\circ + A) - \cot(60^\circ - A) = 3\cot 3A

    L.H.S. =1tanA+1tan(60+A)1tan(60A)= \frac{1}{\tan A} + \frac{1}{\tan(60^\circ + A)} - \frac{1}{\tan(60^\circ - A)}

    =1tanA+13tanA3+tanA1+3tanA3tanA= \frac{1}{\tan A} + \frac{1 - \sqrt{3}\tan A}{\sqrt{3} + \tan A} - \frac{1 + \sqrt{3}\tan A}{\sqrt{3} - \tan A}

    =1tanA8tanA3tan2A=3(13tan2A)3tanAtan3A=3tan3A= \frac{1}{\tan A} - \frac{8\tan A}{3 - \tan^2A} = \frac{3(1 - 3\tan^2A)}{3\tan A - \tan^3A} = \frac{3}{\tan 3A}

    =3cot3A== 3\cot 3A = R.H.S.

  35. We have to prove that cos4A=18cos2A+8cos4A\cos 4A = 1 - 8\cos^2A + 8\cos^4A

    L.H.S. =cos4A=2cos22A1=2(2cos2A1)21= \cos 4A = 2\cos^22A - 1 = 2(2\cos^2A - 1)^2 - 1

    =2(4cos4A4cos2A+1)1=2(4\cos^4A - 4\cos^2A + 1) - 1

    =18cos2A+8cos4A== 1 - 8\cos^2A + 8\cos^4A = R.H.S.

  36. We have to prove that sin4A=4sinAcos3A4cosAsin3A\sin 4A = 4\sin A\cos^3A - 4\cos A\sin^3A

    L.H.S. =2sin2Acos2A=4sinAcosA(cos2Asin2A)= 2\sin 2A\cos 2A = 4\sin A\cos A(\cos^2A - \sin^2A)

    =4sinAcos3A4cosAsin3A== 4\sin A\cos^3A - 4\cos A\sin^3A = R.H.S.

  37. We have to prove that cos6A=32cos6A48cos4A+18cos2A1\cos 6A = 32\cos^6A - 48\cos^4A + 18\cos^2A - 1

    L.H.S. =cos6A=(cos23Asin23A)=(4cos3A3cosA)2(3sinA4sin3A)2= \cos 6A = (\cos^23A - \sin^23A) = (4\cos^3A - 3\cos A)^2 - (3\sin A - 4\sin^3A)^2

    =16cos6A+9cos2A24cos4A9sin2A16sin6A+24sin4A= 16\cos^6A + 9\cos^2A -24\cos^4A - 9\sin^2A - 16\sin^6A + 24\sin^4A

    =16cos6A+9cos2A24cos4A9(1cos2A)16(1cos2A)3+24(1cos2A)2= 16\cos^6A + 9\cos^2A -24\cos^4A - 9(1 - \cos^2A) - 16(1 - \cos^2A)^3 + 24(1 - \cos^2A)^2

    =32cos6A48cos4A+18cos2A1== 32\cos^6A - 48\cos^4A + 18\cos^2A - 1 = R.H.S.

  38. We have to prove that tan3Atan2AtanA=tan3Atan2AtanA\tan 3A\tan 2A\tan A = \tan 3A - \tan 2A - \tan A

    Rewriting this as following:

    tanA+tan2A=tan3A(1tanAtan2A)tanA+tan2A1tanAtan2A=tan3A\tan A + \tan 2A = \tan 3A(1 - \tan A\tan 2A)\Rightarrow \frac{\tan A + \tan 2A}{1 - \tan A\tan 2A} = \tan 3A

    tan(A+2A)=tan3A\Rightarrow \tan (A + 2A) = \tan 3A

    Hence, proved.

  39. We have to prove that 2cos2nA+12cosA+1=(2cosA1)(2cos2A1)(2cos22A1)(2cos2n11)\frac{2\cos2^nA + 1}{2\cos A + 1} = (2\cos A - 1)(2\cos 2A - 1)(2\cos2^2A - 1)\ldots(2\cos2^{n - 1} - 1)

    L.H.S. =2cos2nA+12cosA+1= \frac{2\cos2^nA + 1}{2\cos A + 1}

    Multiplying and dividing by 2cosA12\cos A - 1

    =(2cosA1)2cos2nA+14cos2A1=(2cosA1)2cos2nA+12cos2A+1= (2\cos A - 1)\frac{2\cos2^nA + 1}{4\cos^2A - 1} = (2\cos A - 1)\frac{2\cos2^nA + 1}{2\cos2A + 1}

    Multiplying and dividing by 2cos2A12\cos2A - 1

    =(2cosA1)(2cos2A1)2cos2nA+14cos22A1= (2\cos A - 1)(2\cos2A - 1)\frac{2\cos2^nA + 1}{4\cos^22A - 1}

    =(2cosA1)(2cos2A1)2cos2nA+12cos22A+1= (2\cos A - 1)(2\cos2A - 1)\frac{2\cos2^nA + 1}{2\cos2^2A + 1}

    Proceeding similarly we obtain the R.H.S.

  40. Given tanA=17,sinB=110\tan A= \frac{1}{7}, \sin B = \frac{1}{\sqrt{10}}

    cosB=310,tanB=13\therefore \cos B = \frac{3}{\sqrt{10}}, \tan B = \frac{1}{3}

    tan(A+2B)=tanA+tan2B1tanAtan2B\tan(A + 2B) = \frac{\tan A + \tan 2B}{1 - \tan A\tan 2B}

    =tanA+2tanB1tan2B1tanA.2tanB1tan2B= \frac{\tan A + \frac{2\tan B}{1 - \tan^2B}}{1 - \tan A.\frac{2\tan B}{1 - \tan^2B}}

    =17+213119117.213119= \frac{\frac{1}{7} + \frac{2\frac{1}{3}}{1 - \frac{1}{9}}}{1 - \frac{1}{7}.\frac{2\frac{1}{3}}{1 - \frac{1}{9}}}

    =1A+2B=π4= 1 \therefore A + 2B = \frac{\pi}{4}

  41. We have to prove that tan(π4+A)+tan(π4A)=2sec2A\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right) = 2\sec2A

    L.H.S. =1+tanA1tanA+1tanA1+tanA= \frac{1 + \tan A}{1 - \tan A} + \frac{1 - \tan A}{1 + \tan A}

    =(1+tanA)2+(1tanA)21tan2A=2+2tan2A1tan2A= \frac{(1 + \tan A)^2 + (1 - \tan A)^2}{1 - \tan^2A} = \frac{2 + 2\tan^2A}{1 - \tan^2A}

    =2(sin2A+cos2A)cos2Asin2A=2cos2A=2sec2A== \frac{2(\sin^2A + \cos^2A)}{\cos^2A - \sin^2A} = \frac{2}{\cos 2A} = 2\sec 2A = R.H.S.

  42. We have to prove that 3cosec20sec20=4\sqrt{3}\cosec 20^\circ - \sec 20^\circ = 4

    L.H.S. =3sin201cos20= \frac{\sqrt{3}}{\sin20^\circ} - \frac{1}{\cos20^\circ}

    =4(32)cos2012sin202sin20cos20= \frac{4(\frac{\sqrt{3}}{2})\cos20^\circ - \frac{1}{2}\sin20^\circ}{2\sin20^\circ\cos^20\circ}

    =4(sin(5020))sin40=4== \frac{4(\sin(50^\circ - 20^\circ))}{\sin40^\circ} = 4 = R.H.S.

  43. We have to prove that tanA+2tan2A+4tan4A+8cot8A=cotA\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A

    tanAcotA=sin2Acos2AsinAcosA=2cos2Asin2A=2cot2A\tan A - \cot A = \frac{\sin^2A - \cos^2A}{\sin A\cos A} = -\frac{2\cos 2A}{\sin 2A} = -2\cot 2A

    Similarly, 2tan2A2cot2A=4cot4A2\tan 2A - 2\cot 2A = -4\cot 4A

    and 4tan4A4cot4A=8cot8A4\tan 4A - 4\cot 4A = -8\cot 8A

    Thus, tanA+2tan2A+4tan4A+8cot8A=cotA\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A

  44. We have to prove that cos2A+cos2(2π3A)+cos2(2π3+A)=32\cos^2A + \cos^2\left(\frac{2\pi}{3} - A\right) + \cos^2\left(\frac{2\pi}{3} + A\right) = \frac{3}{2}

    2cos2A+2cos2(2π3A)+2cos2(2π3+A)=3\Rightarrow 2\cos^2A + 2\cos^2\left(\frac{2\pi}{3} - A\right) + 2\cos^2\left(\frac{2\pi}{3} + A\right) = 3

    L.H.S. =cos2A+1+cos(4π32A)+1+cos(4π3+2A)+1= \cos 2A + 1 + \cos\left(\frac{4\pi}{3} - 2A\right) + 1 + \cos\left(\frac{4\pi}{3} + 2A\right) + 1

    =3+cos2A+2cos(4π3)cos2A=3== 3 + \cos2A + 2\cos\left(\frac{4\pi}{3}\right)\cos2A = 3 = R.H.S.

  45. 2sin2A+4cos(A+B)sinAsinB+cos2(A+B)2\sin^2A + 4\cos (A + B)\sin A\sin B + \cos2(A + B)

    =2sin2A+2cos(A+B)2sinAsinB+cos2(A+B)= 2\sin^2A + 2\cos(A + B)2\sin A\sin B + \cos2(A + B)

    =2sin2A+2cos(A+B)[cos(AB)cos(A+B)]+cos2(A+B)= 2\sin^2A + 2\cos(A + B)[\cos(A - B) - \cos(A + B)] + \cos2(A + B)

    =2sin2A+2cos(A+B)cos(AB)2cos2(A+B)+cos2(A+B)= 2\sin^2A + 2\cos(A + B)\cos(A - B) - 2\cos^2(A + B) + \cos2(A + B)

    =2sin2A+2(cos2Asin2B)2cos2(A+B)+2cos2(A+B)1= 2\sin^2A + 2(\cos^2A - \sin^2B) - 2\cos^2(A + B) + 2\cos^2(A + B) - 1

    =2(sin2A+cos2A)2sin2B1=12sin2B= 2(\sin^2A + \cos^2A) -2\sin^2B - 1 = 1 -2\sin^2B which is independent of AA

  46. Given, cosA=12(a+1a)\cos A = \frac{1}{2}\left(a + \frac{1}{a}\right)

    cos2A=2cos2A1=2.14(a+1a)21\cos 2A = 2\cos^2A - 1 = 2.\frac{1}{4}\left(a + \frac{1}{a}\right)^2 - 1

    =12(a2+1a2)= \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)

  47. We have to prove that cos2A+sin2Acos2B=cos2B+sin2Bcos2A\cos^2A + \sin^2A\cos 2B = \cos^2B + \sin^2B\cos 2A

    cos2Acos2B=sin2Bcos2Asin2Acos2B\Rightarrow \cos^2A - \cos^2B = \sin^2B\cos2A - \sin^2A\cos2B

    R.H.S. =sin2Bcos2Asin2Acos2B= \sin^2B\cos2A - \sin^2A\cos2B

    =sin2B(cos2Asin2A)sin2A(cos2Bsin2B)= \sin^2B(\cos^2A - \sin^2A) - \sin^2A(\cos^2B - \sin^2B)

    =cos2Asin2Bsin2Acos2B=cos2A(1cos2B)(1cos2A)cos2B= \cos^2A\sin^2B - \sin^2A\cos^2B = \cos^2A(1 - \cos^2B) - (1 - \cos^2A)\cos^2B

    =cos2Acos2B== \cos^2A - \cos^2B = R.H.S.

  48. We have to prove that 1+tanAtan2A=sec2A1 + \tan A\tan 2A = \sec 2A

    L.H.S. =1+tanAtan2A=1+tanA.2tanA1tan2A= 1 + \tan A\tan 2A = 1 + \tan A.\frac{2\tan A}{1 - \tan^2A}

    =1+tan2A1tan2A=cos2A+sin2Acos2Asin2A= \frac{1 + \tan^2A}{1 - \tan^2A} = \frac{\cos^2A + \sin^2A}{\cos^2A - \sin^2A}

    =1cos2A=sec2A== \frac{1}{\cos 2A} = \sec 2A = R.H.S.

  49. We have to prove that 1+sin2A1sin2A=(1+tanA1tanA)2\frac{1 + \sin 2A}{1 - \sin 2A} = \left(\frac{1 + \tan A}{1 - \tan A}\right)^2

    L.H.S. =1+sin2A1sin2A=sin2A+cos2A+2sinAcosAsin2A+cos2A2sinAcosA= \frac{1 + \sin 2A}{1 - \sin 2A} = \frac{\sin^2A + \cos^2A + 2\sin A\cos A}{\sin^2A + \cos^2A - 2\sin A\cos A}

    =(sinA+cosAsinAcosA)2= \left(\frac{\sin A + \cos A}{\sin A - \cos A}\right)^2

    Dividing numerator and denominator by cos2A,\cos^2A, we get

    =(1+tanA1tanA)2== \left(\frac{1 + \tan A}{1 - \tan A}\right)^2 = R.H.S.

  50. We have to prove that 1sin103cos10=4\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} = 4