5. Trigonometrical Ratios Solutions#

Go through the list of relationship between trigonometrical ratios as the problems are centered around those.

  1. L.H.S. =1cosA1+cosA= \sqrt{\frac{1- \cos A}{1 + \cos A}}

    Multiplying and dividing with 1cosA,1 - \cos A, we get

    =(1cosA)21cos2A= \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}}

    =(1cosAsinA)2= \sqrt{\left(\frac{1 - \cos A}{\sin A}\right)^2}

    =1cosAsinA=cosecAcotA== \frac{1 - \cos A}{\sin A} = \cosec A - \cot A = R.H.S.

  2. L.H.S. 1+tan2A+1+cot2A\sqrt{1 + \tan^2A + 1 + \cot^2A} [sec2A=1+tan2A\because \sec^2A = 1 + \tan^2A and cosec2A=1+cot2A\cosec^2A = 1 + \cot^2 A ]

    =tan2A+2tanAcotA+cot2A= \sqrt{\tan^2A + 2\tan A\cot A + \cot^2A} [tanAcotA=1\tan A\cot A = 1 ]

    =(tanA+cotA)2== \sqrt{(\tan A + \cot A)^2} = R.H.S.

  3. L.H.S =(1sin2AsinA)(1cos2AcosA)(sin2A+cos2AsinAcosA)= \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right)\left(\frac{\sin^2A + \cos^2A}{\sin A \cos A}\right)

    =cos2AsinAsin2AcosA1sinAcosA=1= \frac{\cos^2A}{\sin A}\frac{\sin^2 A}{\cos A}\frac{1}{\sin A \cos A} = 1

  4. L.H.S. =cos4Asin4A+1=(cos2A+sin2A)(cos2Asin2A)+1= \cos^4A - \sin^4A + 1= (\cos^2A + \sin^2A)(\cos^2A - \sin^2A) + 1

    =cos2Asin2A+1=2cos2A= \cos^2A - \sin^2A + 1 = 2\cos^2A

  5. L.H.S. =(sinA+cosA)(1sinAcosA)=(sinA+cosA)(sin2A+cos2AsinAcosA)= (\sin A + \cos A)(1 - \sin A\cos A) = (\sin A + \cos A)(\sin^2A + \cos^2A - \sin A\cos A)

    =sin3A+cos3A== \sin^3A + \cos^3A = R.H.S.

  6. L.H.S =sinA1+cosA+1+cosAsinA=sin2A+(1+cosA)2sinA(1+cosA)= \frac{\sin A}{1 + \cos A}+\frac{1 + \cos A}{\sin A} = \frac{\sin^2A + (1 + \cos A)^2}{\sin A(1 + \cos A)}

    =sin2A+1+2cosA+cos2AsinA(1+cosA)=1+1+2cosA(sinA)(1+cosA))= \frac{\sin^2A + 1 + 2\cos A + \cos^2A}{\sin A(1 + \cos A)} = \frac{1 + 1 + 2\cos A}{(\sin A)(1 + \cos A))}

    =2sinA=2cosecA== \frac{2}{\sin A} = 2\cosec A = R.H.S.

  7. L.H.S. =sin6Acos6A=(sin2A+cos2A)33cos4Asin2A3cos2Asin4A= \sin^6A - cos^6A = (\sin^2A + \cos^2A)^3 - 3\cos^4A\sin^2A - 3\cos^2A\sin^4A

    =13sin2Acos2A(cos2A+sin2A)=13sin2Acos2A== 1 - 3\sin^2A\cos^2A(\cos^2A + \sin^2A) = 1 - 3\sin^2A\cos^2A = R.H.S.

  8. L.H.S. =1sinA1+sinA= \sqrt{\frac{1 - \sin A}{1 + \sin A}}

    Multiplying and dividing with 1sinA,1 - \sin A, we get

    =(1sinA)21sin2A=(1sinA)2cos2A=(1sinAcosA)2= \sqrt{\frac{(1 - \sin A)^2}{1 - \sin^2A}} = \sqrt{\frac{(1 - \sin A)^2}{\cos^2A}} = \sqrt{\left(\frac{1 - \sin A}{\cos A}\right)^2}

    =1sinAcosA=secAtanA= \frac{1 - \sin A}{\cos A} = \sec A - \tan A

  9. L.H.S. =cosecAcosecA1+cosecAcosecA+1=1sinA1sinA1+1sinA1sinA+1= \frac{\cosec A}{\cosec A - 1} + \frac{\cosec A}{\cosec A + 1} = \frac{\frac{1}{\sin A}}{\frac{1}{\sin A} - 1} + \frac{\frac{1}{\sin A}}{\frac{1}{\sin A} + 1}

    =11sinA+11+sinA=1(1sinA)(1+sinA)=1cos2A=sec2A== \frac{1}{1 - \sin A} + \frac{1}{1 + \sin A} = \frac{1}{(1 - \sin A)(1 + \sin A)} = \frac{1}{\cos^2A} = \sec^2A = R.H.S.

  10. L.H.S. =cosecAtanA+cotA=1sinAsinAcosA+cosAsinA= \frac{\cosec A}{\tan A + \cot A} = \frac{\frac{1}{\sin A}}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}

    =1sinAsin2A+cos2AsinAcosA=1sinA1sinAcosA=cosA== \frac{\frac{1}{\sin A}}{\frac{\sin^2A + \cos^2A}{\sin A\cos A}} = \frac{\frac{1}{\sin A}}{\frac{1}{\sin A\cos A}} = \cos A = R.H.S.

  11. L.H.S. =(secA+cosA)(secAcosA)=sec2Acos2A=1+tan2A1+sin2A=tan2A+sin2A== (\sec A + \cos A)(\sec A - \cos A) = \sec^2A - \cos^2A = 1 + \tan^2A - 1 + \sin^2A = \tan^2A + \sin^2A = R.H.S.

  12. L.H.S. =1tanA+cotA=1sinAcosA+cosAsinA=1sin2A+cos2AsinAcosA=sinAcosA== \frac{1}{\tan A + \cot A} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2A + \cos^2A}{\sin A\cos A}} = \sin A\cos A = R.H.S.

  13. L.H.S =1tanA1+tanA=11cotA1+1cotA=cotA1cotA+1== \frac{1 - \tan A}{1 + \tan A} = \frac{1 - \frac{1}{\cot A}}{1 + \frac{1}{\cot A}} = \frac{\cot A - 1}{\cot A + 1} = R.H.S.

  14. L.H.S. =1+tan2A1+cot2A=1+sin2Acos2A1+cos2AsinA= \frac{1 + \tan^2A}{1 + \cot^2A} = \frac{1 + \frac{\sin^2A}{\cos^2A}}{1 + \frac{\cos^2A}{\sin^A}}

    =cos2A+sin2Acos2Asin2A+cos2Asin2A=sin2Acos2A== \frac{\frac{\cos^2A + \sin^2A}{\cos^2A}}{\frac{\sin^2A + \cos^2A}{\sin^2A}} = \frac{\sin^2A}{\cos^2A} = R.H.S.

  15. L.H.S. =secAtanAsecA+tanA= \frac{\sec A - \tan A}{\sec A + \tan A}

    Multiplying and dividing with secAtanA,\sec A - \tan A, we get

    (secAtanA)2sec2Atan2A=sec2A2tanAsecA+tanA=1+tan2A2tanAsecA+tan2A\frac{(\sec A - \tan A)^2}{\sec^2A - \tan^2A} = \sec^2A - 2\tan A\sec A + \tan^A = 1 + \tan^2A - 2\tan A\sec A + \tan^2A

    =12tanAsecA+2tan2A== 1 - 2\tan A\sec A + 2\tan^2A = R.H.S.

  16. L.H.S. =1secAtanA= \frac{1}{\sec A - \tan A}

    Multiplying and dividing with secA+tanA,\sec A + \tan A, we get

    =secA+tanAsec2A=tan2A=secA+tanA== \frac{\sec A + \tan A}{\sec^2A = \tan^2A} = \sec A + \tan A = R.H.S.

  17. L.H.S. =tanA1cotA+cotA1tanA= \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}

    =sinAcosA1cosAsinA+cosAsinA1sinAcosA= \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}

    =sin2AcosA(sinAcosA)+cos2AsinA(cosAsinA)= \frac{\sin^2A}{\cos A(\sin A - \cos A)} + \frac{\cos^2A}{\sin A(\cos A - \sin A)}

    =sin3Acos3AsinAcosA(sinAcosA)=sin2A+cos2A+sinAcosAsinAcosA= \frac{\sin^3A - \cos^3A}{\sin A\cos A(\sin A - \cos A)} = \frac{\sin^2A + \cos^2A + \sin A\cos A}{\sin A\cos A}

    =1+sinAcosAsinAcosA=1+cosecAsecA== \frac{1 + \sin A \cos A}{\sin A\cos A} = 1 + \cosec A\sec A = R.H.S.

  18. L.H.S. =cosA1tanA+sinA1cotA=cosA1sinAcosA+sinA1cosAsinA= \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}}

    =cos2AcosAsinA+sin2AsinAcosA=cos2Asin2AcosAsinA= \frac{\cos^2A}{\cos A - \sin A} + \frac{\sin^2A}{\sin A - \cos A} = \frac{\cos^2A - \sin^2A}{\cos A - \sin A}

    =cosA+sinA== \cos A + \sin A = R.H.S.

  19. L.H.S. =(sinA+cosA)(tanA+cotA)=(sinA+cosA)(sinAcosA+cosAsinA)= (\sin A + \cos A)(\tan A + \cot A) = (\sin A + \cos A)(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})

    =(sinA+cosA)sin2A+cos2AsinAcosA=sinA+cosAsinAcosA=secA+cosecA== (\sin A + \cos A)\frac{\sin^2A + \cos^2A}{\sin A\cos A} = \frac{\sin A + \cos A}{\sin A\cos A} = \sec A + \cosec A = R.H.S.

  20. L.H.S. =sec4Asec2A=(1+tan2A)1tan2A=1+2tan2A+tan4A1tan2A=tan4A+tan2A== \sec^4A - \sec^2A = (1 + \tan^2A) - 1 - \tan^2A = 1 + 2\tan^2A + \tan^4A - 1 - \tan^2A = \tan^4A + \tan^2A = R.H.S.

  21. L.H.S. =cot4A+cot2A=(cosec2A1)1+cosec2A1=cosec4A2cosec2A+1+cosec2A1=cosec4Acosec2A== \cot^4A + \cot^2A = (\cosec^2A - 1)^1 + \cosec^2A - 1 = \cosec^4A - 2\cosec^2A + 1 + \cosec^2A -1 = \cosec^4A - \cosec^2A = R.H.S.

  22. L.H.S. =cosec2A1=cot2A=cotA=cosAsinA=cosAcosecA== \sqrt{\cosec^2A - 1} = \sqrt{\cot^2A} = \cot A = \frac{\cos A}{\sin A} = \cos A \cosec A = R.H.S.

  23. L.H.S. =sec2Acosec2A=(1+tan2A)(1+cot2A)=1+tan2A+cot2A+tan2Acos2A=2+tan2A+cot2A== \sec^2A\cosec^2A = (1 + \tan^2A)(1 + \cot^2A) = 1 + \tan^2A + \cot^2A + \tan^2A\cos^2A = 2 + \tan^2A + \cot^2A = R.H.S.

  24. L.H.S. =tan2Asin2A=sin2Acos2Asin2A=sin2Asin2Acos2Acos2A=sin2A(1cos2A)sec2A=sin4Asec2A== \tan^2A - \sin^2A = \frac{\sin^2A}{\cos^2A} - \sin^2A = \frac{\sin^2A - \sin^2A\cos^2A}{\cos^2A} = \sin^2A(1 - \cos^2A)\sec^2A = \sin^4A\sec^2A = R.H.S.

  25. L.H.S. =(1+cotAcosecA)(1+tanA+secA)=(1+cosAsinA1sinA)(1+sinAcosA+1cosA)= (1 + \cot A - \cosec A)(1 + \tan A + \sec A) = \left(1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A}\right)\left(1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A}\right)

    =(sinA+cosA1)(sinA+cosA+1)sinAcosA=(sinA+cosA)21sinAcosA= \frac{(\sin A + \cos A - 1)(\sin A + \cos A + 1)}{\sin A\cos A} = \frac{(\sin A + \cos A)^2 - 1}{\sin A\cos A}

    =sin2A+cos2A+sinAcosA1sinAcosA=1+2sinAcosA1sinAcosA=2== \frac{\sin^2A + \cos^2A + \sin A\cos A - 1 }{\sin A\cos A} = \frac{1 + 2\sin A\cos A - 1}{\sin A\cos A} = 2= R.H.S.

  26. L.H.S. =cotAcosAcotA+cosA=cos2AsinAcosAsinA+cosA= \frac{\cot A\cos A}{\cot A + \cos A} = \frac{\frac{\cos^2 A}{\sin A}}{\frac{\cos A}{\sin A} + \cos A}

    =cos2AsinAsinAcosA(1+sinA)=cosA1+sinA= \frac{\cos^2A}{\sin A}{\frac{\sin A}{\cos A(1 + \sin A)}} = \frac{\cos A}{1 + \sin A}

    Proceeding in same way for R.H.S. we obtain 1sinAcosA\frac{1 - \sin A}{\cos A}

    cosA1+sinA=cosA(1sinA)1sin2A=1sinAcosA=\frac{\cos A}{1 + \sin A} = \frac{\cos A(1 - \sin A)}{1 - \sin^2A} = \frac{1 - \sin A}{\cos A} = R.H.S.

  27. L.H.S. =cotA+tanBcotB+tanA=1tanA+tanB1tanB+tanA= \frac{\cot A + \tan B}{\cot B + \tan A} = \frac{\frac{1}{\tan A} + \tan B}{\frac{1}{\tan B} + \tan A}

    1+tanAtanBtanA1+tanAtanBtanB=tanBtanA=cotAtanB=\frac{\frac{1 + \tan A\tan B}{\tan A}}{\frac{1 + \tan A\tan B}{\tan B}} = \frac{\tan B}{\tan A} = \cot A\tan B = R.H.S.

  28. L.H.S. =(1sec2Acos2A+1cosec2Asin2A)cos2Asin2A= \left(\frac{1}{\sec^2 A - \cos^2A} + \frac{1}{\cosec^2A - \sin^2A}\right)\cos^2A\sin^2A

    =(11cos2Acos2A+11sin2Asin2A)cos2Asin2A= \left(\frac{1}{\frac{1}{\cos^2A} - \cos^2A} + \frac{1}{\frac{1}{\sin^2A} - \sin^2A}\right)\cos^2A\sin^2A

    =(cos2A1cos4A+sin2A1sin4)cos2Asin2A= \left(\frac{\cos^2A}{1 - \cos^4A} + \frac{\sin^2A}{1 - \sin^4}\right)\cos^2A\sin^2A

    =(cos2A(1cos2A)(1+cos2A)+sin2A(1sin2A)(1+sin2A))cos2Asin2A= \left(\frac{\cos^2A}{(1 - \cos^2A)(1 + \cos^2A)} + \frac{\sin^2A}{(1 - \sin^2A)(1 + \sin^2A)}\right)\cos^2A\sin^2A

    =(cos2Asin2A(1+cos2A)+sin2Acos2A(1+sin2A))cos2Asin2A= \left(\frac{\cos^2A}{\sin^2A(1 + \cos^2A)} + \frac{\sin^2A}{\cos^2A(1 + \sin^2A)}\right)\cos^2A\sin^2A

    =(cos4A(1+sin2A)+sin4A(1+cos2A)sin2Acos2A(1+cos2A))(1+sin2A))cos2Asin2A= \left(\frac{\cos^4A(1 + \sin^2A) + \sin^4A(1 + \cos^2A)}{\sin^2A\cos^2A(1 + \cos^2A))(1 + \sin^2A)}\right)\cos^2A\sin^2A

    =(cos4A+sin4A+cos2Asin2A(cos2A+sin2A)(1+cos2A)(1+sin2A))= \left(\frac{\cos^4A + \sin^4A + \cos^2A\sin^2A(\cos^2A + \sin^2A)}{(1 +\cos^2A)(1 + \sin^2A)}\right)

    =((cos2A+sin2A)2cos2Asin2A1+sin2A+cos2A+sin2Acos2A)= \left(\frac{(\cos^2A + \sin^2A)^2 - \cos^2A\sin^2A}{1 + \sin^2A + \cos^2A + \sin^2A\cos^2A}\right)

    =1sin2Acos2A2+sin2Acos2A== \frac{1 - \sin^2A\cos^2A}{2 + \sin^2A\cos^2A} = R.H.S.

  29. L.H.S. =sin8Acos8A=(sin4Acos4A)(sin4A+cos4A)= \sin^8A - \cos^8A = (\sin^4A - \cos^4A)(\sin^4A + \cos^4A)

    =(sin2A+cos2A)(sin2Acos2A)((sin2A+cos2A)22sin2Acos2A)= (\sin^2A + \cos^2A)(\sin^2A - \cos^2A)((\sin^2A + \cos^2A)^2 - 2\sin^2A\cos^2A)

    =(sin2Acos2A)(12sin2Acos2A)== (\sin^2A - \cos^2A)(1 - 2\sin^2A\cos^2A) = R.H.S.

  30. L.H.S. =cosAcosecAsinAsecAcosA+sinA=cosAsinAsinAcosAcosA+sinA= \frac{\cos A\cosec A - \sin A\sec A}{\cos A + \sin A} = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\cos A}}{\cos A + \sin A}

    =cos2Asin2AsinAcosA(cosA+sinA)=cosAsinAsinAcosA=cosecAsecA== \frac{\cos^2A - \sin^2A}{\sin A\cos A(\cos A + \sin A)} = \frac{\cos A \sin A}{\sin A\cos A} = \cosec A - \sec A = R.H.S.

  31. Given, 1cosecAcotA1sinA=1sinA1cosecA+cotA\frac{1}{\cosec A - \cot A} - \frac{1}{\sin A} = \frac{1}{\sin A} - \frac{1}{\cosec A + \cot A}

    Therefore alternatively we can prove that 1cosecAcotA+1cosecA+cotA=2sinA\frac{1}{\cosec A - \cot A} + \frac{1}{\cosec A + \cot A} = \frac{2}{\sin A}

    L.H.S. =11sinAcosAsinA+11sinA+cosAsinA= \frac{1}{\frac{1}{\sin A} - \frac{\cos A}{\sin A}} + \frac{1}{\frac{1}{\sin A} + \frac{\cos A}{\sin A}}

    =sinA1cosAsinA1+cosA=sinA(1+cosA+1cosA)1cos2A=2sinAsin2A=2sinA== \frac{\sin A}{1 - \cos A} - \frac{\sin A}{1 + \cos A} = \frac{\sin A(1 + \cos A + 1 - \cos A)}{1 - \cos^2 A} = \frac{2\sin A}{\sin^2A} = \frac{2}{\sin A} = R.H.S.

  32. L.H.S. =tanA+secA1tanAsecA+1=sinA+1cosAcosAsinA+cosA1)cosA= \frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{\frac{\sin A + 1 - \cos A}{\cos A}}{\frac{\sin A + \cos A - 1)}{\cos A}}

    =1+sinAcosAsinA+cosA1=(1+sinAcosA)(sinA+cosA1)(sinA+cosA1)2= \frac{1 + \sin A - \cos A}{\sin A + \cos A - 1} = \frac{(1 + \sin A - \cos A)(\sin A + \cos A - 1)}{(\sin A + \cos A - 1)^2}

    Solving this yields =1+sinAcosA== \frac{1 + \sin A}{\cos A} = R.H.S.

  33. L.H.S. =(tanA+cosecB)2(cotBsecA)2=tan2A+cosec2B+2tanAcosecBcot2Bsec2A+2secAcotB= (\tan A + \cosec B)^2 - (\cot B - \sec A)^2 = \tan^2A + \cosec^2B + 2\tan A\cosec B - \cot^2B - \sec^2A+ 2\sec A\cot B

    =(cosec2Bcot2B)(sec2Atan2A)+2tanAcotB(cosecBcotB+secAtanA)= (\cosec^2B - \cot^2B) - (sec^2A - tan^2A) + 2\tan A\cot B\left(\frac{\cosec B}{\cot B} + \frac{\sec A}{\tan A}\right)

    =11+2tanAcotB(secB+cosecA)== 1 - 1 + 2\tan A\cot B\left(\sec B + \cosec A\right) = R.H.S.

  34. L.H.S. =2sec2Asec4A2cosec2A+cosec4A=2(1+tan2A)(1+tan2A)22(1+cot2A)+(1+cot2A)2= 2\sec^2 A - \sec^4A - 2\cosec^2A + \cosec^4A = 2(1 + \tan^2A) - (1 + \tan^2A)^2 - 2(1 + \cot^2A) + (1 + \cot^2A)^2

    =2+2tan2A12tan2A+tan4A22cot2A+1+2cot2A+cot4A=cot4Atan4A== 2 + 2\tan^2A - 1 - 2\tan^2A + \tan^4A - 2 - 2\cot^2A + 1 + 2\cot^2A + \cot^4A = \cot^4A - \tan^4A = R.H.S.

  35. L.H.S. =(sinA+cosecA)2+(cosA+secA)2=sin2A+cosec2A+2sinAcosecA+cos2A+sec2A+2secAcosecA= (\sin A + \cosec A)^2 + (\cos A + \sec A)^2 = \sin^2A + \cosec^2A + 2\sin A\cosec A + \cos^2A + \sec^2A + 2\sec A\cosec A

    =(sin2A+cos2A)+4+cosec2A+sec2A=1+4+1+cot2A+1+tan2A=cot2A+tan2A+7== (\sin^2A + \cos^2A) + 4 + \cosec^2A + \sec^2A = 1 + 4 + 1 + \cot^2A + 1 + \tan^2A = \cot^2A + \tan^2A + 7 = R.H.S.

  36. L.H.S =(cosecA+cotA)(1sinA)(secA+tanA)(1cosA)= (\cosec A + \cot A)(1 - \sin A) - (\sec A + \tan A)(1 - \cos A)

    =(1sinA+cosAsinA)(1sinA)(1cosA+sinAcosA)(1cosA)= \left(\frac{1}{\sin A} + \frac{\cos A}{\sin A}\right)(1 - \sin A) - \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 - \cos A)

    =(1+cosA)(1sinA)sinA(1+sinA)(1cosA)cosA= \frac{(1 + \cos A)(1 - \sin A)}{\sin A} - \frac{(1 + \sin A)(1 - \cos A)}{\cos A}

    =cosA(1+cosAsinAcosAsinA)sinA(1+sinAcosAsinAcosA)sinAcosA= \frac{\cos A(1 + \cos A - \sin A - \cos A\sin A) -\sin A(1 + \sin A - \cos A - \sin A\cos A)}{\sin A\cos A}

    =(1sinAcosA)(cosAsinA)+(cosAsinA)(cosA+sinA)sinAcosA= \frac{(1 - \sin A\cos A)(\cos A - \sin A) + (\cos A - \sin A)(\cos A + \sin A)}{\sin A\cos A}

    =(cosAsinA)(1cosAsinA+cosA+sinA)sinAcosA= \frac{(\cos A - \sin A)(1 - \cos A\sin A + \cos A + \sin A)}{\sin A\cos A}

    =((cosAsinA)sinAcosA)(21cosAsinA+cosA+sinA)= \left(\frac{(\cos A - \sin A)}{\sin A\cos A}\right)(2 - 1 - \cos A\sin A + \cos A + \sin A)

    =(cosecAsecA)(2(1cosA)(1sinA))== (\cosec A - \sec A)(2 - (1 - \cos A)(1 - \sin A)) = R.H.S.

  37. L.H.S =(1+cotA+tanA)(sinAcosA)=(1+cosecAsecA+secAcosecA)(1cosecA1secA)= (1 + \cot A + \tan A)(\sin A - \cos A) = \left(1 + \frac{\cosec A}{\sec A} + \frac{\sec A}{\cosec A}\right)\left(\frac{1}{\cosec A} - \frac{1}{\sec A}\right)

    =(secAcosecA+secAcosecAsecAcosecA)(secAcosecAsecAcoseca)= \left(\frac{\sec A\cosec A + \sec A | \cosec A}{\sec A\cosec A}\right)\left(\frac{\sec A - \cosec A}{\sec A\cosec a}\right)

    =sec3Acosec3Asec2Acosec2A=secAcosec2AcosecAsec2A== \frac{\sec^3A - \cosec^3A}{\sec^2A\cosec^2A} = \frac{\sec A}{\cosec^2A} - \frac{\cosec A}{\sec^2A} = R.H.S.

  38. Given, 1secAtanA1cosA=1cosA1secA+tanA\frac{1}{\sec A - \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A + \tan A}

    So alternatively we can prove that 1secAtanA+1secA+tanA=2cosA\frac{1}{\sec A - \tan A} + \frac{1}{\sec A + \tan A} = \frac{2}{\cos A}

    L.H.S. =secA+tanA+secAtanAsec2Atan2A=2secA=2cosA== \frac{\sec A + \tan A + \sec A - \tan A}{\sec^2A - tan^2A} = 2\sec A = \frac{2}{\cos A} = R.H.S.

  39. L.H.S. =3(sinAcosA)4+4(sin6A+cos6A)+6(sinA+cosA)2= 3(\sin A - \cos A)^4 + 4(\sin^6 A + \cos^6 A) + 6(\sin A + \cos A)^2

    =3[(sinAcosA)2]2+4[(sin2A)3+(cos2A)3]+6(sin2A+cos2A+2cosAsinA)= 3[(\sin A - \cos A)^2]^2 + 4[(\sin^2A)^3 + (\cos^2A)^3] + 6(\sin^2A + \cos^2A + 2\cos A\sin A)

    =3[(sin2A+cos2A2sinAcosA)]2+4(sin2A+cos2A)(sin4A+cos4Asin2Acos2A)+6(1+2cosAsinA)= 3[(\sin^2A + \cos^2A - 2\sin A\cos A)]^2 + 4(\sin^2A + \cos^2A)(\sin^4A + \cos^4A - \sin^2A\cos^2A) + 6(1 + 2\cos A\sin A)

    =3(12sinAcosA)2+4(sin4A+cos4Asin2Acos2A)+6(1+2cosAsinA)= 3(1 - 2\sin A\cos A)^2 + 4(\sin^4A + \cos^4A - \sin^2A\cos^2A) + 6(1 + 2\cos A\sin A)

    =3(1+4sin2Acos2A4cosAsinA)+4[(sin2A+cos2A)22sin2Acos2Asin2Acos2A]+6(1+2sinAcosA)= 3(1 + 4\sin^2A\cos^2A - 4\cos A\sin A) + 4[(\sin^2A + \cos^2A)^2 - 2\sin^2A\cos^2A - \sin^2A\cos^2A] + 6(1 + 2\sin A\cos A)

    =3+12sin2Acos2A12cosAsinA+4(13sin2Acos2A)+6+12sinAcosA= 3 + 12\sin^2A\cos^2A -12\cos A\sin A + 4(1 - 3\sin^2A\cos^2A) + 6 + 12\sin A\cos A

    =13== 13 = R.H.S.

  40. L.H.S. =1+cosA1cosA= \sqrt{\frac{1 + \cos A}{1 - \cos A}}

    Multiplying and dividing with 1+cosA,1 + \cos A, we get

    =(1+cosA)21cos2A=(1+cosAsinA)2=cosecA+cotA== \sqrt{\frac{(1 + \cos A)^2}{1 - \cos^2A}} = \sqrt{\left(\frac{1 + \cos A}{\sin A}\right)^2} = \cosec A + \cot A = R.H.S.

  41. L.H.S. =cosA1+sinA+cosA1sinA= \frac{\cos A}{1 + \sin A} + \frac{\cos A}{1 - \sin A}

    cosA(1sinA)+cosA(1+sinA)1sin2A\frac{\cos A(1 - \sin A) + \cos A(1 + \sin A)}{1 - \sin^2A}

    =2cosAcos2A=2secA== \frac{2\cos A}{\cos^2A} = 2\sec A = R.H.S.

  42. L.H.S. =tanAsecA1+tanAsecA+1= \frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1}

    =tanAsecA+tanA+tanAsecAtanA(sec2A1)= \frac{\tan A\sec A + \tan A + \tan A\sec A - \tan A}{(\sec^2A - 1)}

    =2tanAsecAtan2A=2cosecA== \frac{2\tan A\sec A}{\tan^2 A} = 2\cosec A = R.H.S.

  43. L.H.S. =11sinA11+sinA= \frac{1}{1 - \sin A} - \frac{1}{1 + \sin A}

    =1+sinA1+sinA1sin2A=2sinAcos2A=2secAtanA== \frac{1 + \sin A - 1 + \sin A}{1 - \sin^2A} = \frac{2\sin A}{\cos^2A} = 2\sec A\tan A = R.H.S.

  44. L.H.S. =1+tan2A1+cot2A=1+sin2Acos2A1+cos2Asin2A= \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{1 + \frac{\sin^2A}{\cos^2A}}{1 + \frac{\cos^2A}{\sin^2A}}

    =sin2A+cos2Acos2Asin2A+cos2Asin2A=sin2Acos2A=tan2A= \frac{\frac{\sin^2A + \cos^2A}{\cos^2A}}{\frac{\sin^2A + \cos^2A}{\sin^2A}} = \frac{\sin^2A}{\cos^2A} = \tan^2A

    R.H.S. =(1tanA1cotA)2=(1sinAcosA1cosAsinA)= \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \left(\frac{1 - \frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}}\right)

    =(cosAsinAcosAsinAcosAsinA)2= \left(\frac{\frac{\cos A - sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}}\right)^2

    =sin2Acos2A=tan2A= \frac{\sin^2A}{\cos^2A} = \tan^2A

    Hence, L.H.S. = R.H.S.

  45. L.H.S. =1+2tan2Acos2A=1+2sin2Acos4A=cos4A+2sin2Acos4A= 1 + \frac{2\tan^2 A}{\cos^2 A} = 1 + \frac{2\sin^2A}{\cos^4A} = \frac{\cos^4A + 2\sin^2A}{\cos^4A}

    =(1sin2A)2+2sin2Acos4A=12sin2A+sin4A+2sin2Acos4A= \frac{(1 - \sin^2A)^2 + 2\sin^2A}{\cos^4A} = \frac{1 - 2\sin^2A + \sin^4A + 2\sin^2A}{\cos^4A}

    =tan4A+sec4A== \tan^4A + sec^4A = R.H.S.

  46. L.H.S. =(1sinAcosA)2=12cosA2sinA+2sinAcosA+sin2A+cos2A= (1 - \sin A - \cos A)^2 = 1 - 2\cos A - 2\sin A + 2\sin A\cos A + \sin^2A + \cos^2A

    =22cosA2sinA+2sinAcosA=2(1cosA)(1sinA)== 2 - 2\cos A - 2\sin A + 2\sin A\cos A = 2(1 - \cos A)(1 - \sin A) = R.H.S.

        1. =cotA+cosecA1cotAcosecA+1= \frac{\cot A + \cosec A - 1}{\cot A - \cosec A + 1}

    =cosA+1sinAcosA+sinA1= \frac{\cos A + 1 - \sin A}{\cos A + \sin A - 1}

    Multiplying and dividing with cosA(1sinA),\cos A - (1 - \sin A), we get

    =cos2A1+2sinAsin2Acos2A+sin2A+12cosA2sinA+2sinAcosA= \frac{\cos^2A - 1 + 2\sin A - \sin^2A}{\cos^2A + \sin^2A + 1 - 2\cos A - 2\sin A + 2\sin A\cos A}

    =2sinA2sin2A2(1cosA)(1sinA)=sinA1cosA= \frac{2\sin A - 2\sin^2A}{2(1 - \cos A)(1 - \sin A)} = \frac{\sin A}{1 - \cos A}

    Multiplyig numerator and denominator with 1+cosA,1 + \cos A, we get

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

  47. L.H.S. =(sinA+secA)2+(cosA+cosecA)2= (\sin A + \sec A)^2 + (\cos A + \cosec A)^2

    =(sinAcosA+1cosA)2+(cosAsinA+1sinA)2= \left(\frac{\sin A\cos A + 1}{\cos A}\right)^2 + \left(\frac{\cos A\sin A + 1}{\sin A}\right)^2

    =(1+sinAcosA)2(1cos2A+1sin2A)= (1 + \sin A\cos A)^2 \left(\frac{1}{\cos^2A} + \frac{1}{\sin^2A}\right)

    =(1+sinAcosA)2(sin2A+cos2Asin2Acos2A)= (1 + \sin A\cos A)^2\left(\frac{\sin^2A + \cos^2A}{\sin^2A\cos^2A}\right)

    =(1+sinAcosAsinAcosA)2=(1+secAcosecA)2== \left(\frac{1 + \sin A\cos A}{\sin A\cos A}\right)^2 = (1 + \sec A\cosec A)^2 = R.H.S.

  48. L.H.S. =2sinAtanA(1tanA)+2sinAsec2A(1+tanA)2= \frac{2\sin A\tan A(1 - \tan A) + 2\sin A\sec^2A}{(1 + \tan A)^2}

    =2sinA(tanAtan2A+sec2A)(1+tanA)2=2sinA(1+tanA)(1+tanA)2= \frac{2\sin A(\tan A - \tan^2A + \sec^2A)}{(1 + \tan A)^2} = \frac{2\sin A(1 + \tan A)}{(1 + \tan A)^2}

    =2sinA1+tanA== \frac{2\sin A}{1 + \tan A} = R.H.S.

  49. Given, 2sinA=2cosA2\sin A = 2 - \cos A

    cosA=22sinA,\cos A = 2 - 2\sin A, squaring both sides we, get cos2A=48sinA+4sin2A\cos^2A = 4 - 8\sin A + 4\sin^2A

    1sin2A=48sinA+4sin2A5sin2A8sinA+3=01 - \sin^2A = 4 - 8\sin A + 4\sin^2A \Rightarrow 5\sin^2A - 8\sin A + 3 = 0

    5sin2A5sinA3sinA+3=0sinA=1,355\sin^2A - 5\sin A - 3\sin A + 3 = 0 \Rightarrow \sin A = 1, \frac{3}{5}

  50. Given 8sinA=4+cosA8sinA4=cosA8\sin A = 4 + \cos A \Rightarrow 8\sin A - 4 = \cos A

    Squaring both sides, we get

    64sin2A64sinA+16=cos2A=1sin2A64\sin^2A - 64\sin A + 16 = \cos^2A = 1 - \sin^2A

    65sin2A64sinA+15=065\sin^2A - 64\sin A + 15 = 0

    65sin2A39sinA25sinA+15=065\sin^2A - 39\sin A - 25\sin A + 15 = 0

    sinA=35,513\sin A = \frac{3}{5}, \frac{5}{13}

  51. Given, tanA+secA=1.5\tan A + \sec A = 1.5

    1+sinA=1.5cosA2+2sinA=3cosA\Rightarrow 1 + \sin A = 1.5 \cos A \Rightarrow 2 + 2\sin A = 3\cos A

    Squaring both sides, we get

    4+8sinA+4sin2A=99sin2A4 + 8\sin A + 4\sin^2A = 9 - 9\sin^2A

    13sin2A+8sinA5=013\sin^2A + 8\sin A - 5 =0

    sinA=1,513\sin A = -1, \frac{5}{13}

  52. Given, cotA+cosecA=51+cosA=5sinA\cot A + \cosec A = 5 \Rightarrow 1 + \cos A = 5\sin A

    Squarig both sides, we get

    1+2cosA+cos2A=25sin2A=25(1cos2A)1 + 2\cos A + \cos^2A = 25\sin^2A = 25(1 - \cos^2A)

    26cos2A+2cosA24=026\cos^2A + 2\cos A - 24 = 0

    26cos2A+26cosA24cosA24=026\cos^2A + 26\cos A - 24\cos A - 24 = 0

    cosA=1,1213\cos A= -1, \frac{12}{13}

  53. 3sec4A+8=10sec2A3(sec2A)2+8=10(1+tan2A)3\sec^4 A + 8 = 10\sec^2A \Rightarrow 3(sec^2A)^2 + 8 = 10(1 + \tan^2A)

    3(1+tan2A)2+8=10+10tan2A\Rightarrow 3(1 + \tan^2A)^2 + 8 = 10 + 10\tan^2A

    3+6tan2A+3tan4A+8=10+10tan2A3 + 6\tan^2A + 3\tan^4A + 8 = 10 + 10\tan^2A

    3tan4A4tan2A+1=03\tan^4A - 4\tan^2A + 1 = 0

    tanA=±1,±13\tan A = \pm1, \pm\frac{1}{\sqrt{3}}

  54. Given, tan2A+secA=5sec2A1+secA=5\tan^2A + \sec A = 5 \Rightarrow \sec^2A - 1 + \sec A = 5

    sec2A+secA6=0\sec^2A + \sec A - 6 = 0

    secA=3,2cosA=13,12\sec A = -3, 2\Rightarrow \cos A = -\frac{1}{3}, \frac{1}{2}

  55. Given tanA+cotA=2tanA+1tanA=2\tan A + \cot A = 2 \Rightarrow \tan A + \frac{1}{\tan A} = 2

    tan2A2tanA+1=0\tan^2A - 2\tan A + 1 = 0

    tanA=1sinA=12\tan A = 1 \Rightarrow \sin A = \frac{1}{\sqrt{2}}

  56. Given, sec2A=2+2tanAtan2A+1=2+2tanA\sec^2A = 2 + 2\tan A \Rightarrow \tan^2A + 1 = 2 + 2\tan A

    tan2A2tanA1=0\tan^2A - 2\tan A - 1 = 0

    tanA=2±4+42=1±2\tan A = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}

  57. Given, tanA=2x(x+1)2x+1\tan A = \frac{2x(x + 1)}{2x + 1}

    sinA=2x(x+1)[2x(x+1)]2+(2x+1)2\sin A = \frac{2x(x + 1)}{\sqrt{[2x(x + 1)]^2 + (2x + 1)^2}}

    cosA=2x+1[2x(x+1)]2+(2x+1)2\cos A = \frac{2x + 1}{\sqrt{[2x(x + 1)]^2 + (2x + 1)^2}}

  58. Given, 3sinA+5cosA=5,3\sin A + 5\cos A = 5, let 5sinA3cosA=x5\sin A - 3\cos A = x

    Squaring and adding, we get

    9sin2A+25cos2A+30sinAcosA+25sin2A+9cos2A30sinAcosA=25+x29\sin^2A + 25\cos^2A + 30\sin A\cos A + 25\sin^2A + 9\cos^2A - 30\sin A\cos A = 25 + x^2

    9(sin2A+cos2A)+25(cos2A+sin2A)=25+x29(\sin^2A + \cos^2A) + 25(\cos^2A + \sin^2A) = 25 + x^2

    34=25+x2x2=9x=±334 = 25 + x^2 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3

  59. Given, secA+tanA=secAtanA\sec A + \tan A = \sec A - \tan A

    Multiplying both sides with secA+tanA,\sec A + \tan A, we get

    (secA+tanA)2=sec2Atan2A=1(\sec A + \tan A)^2 = \sec^2A - \tan^2A = 1

    secA+tanA=±1\sec A + \tan A = \pm 1

    We can prove that secAtanA\sec A - \tan A to be ±1\pm 1 by multiplying given equation with secAtanA\sec A - \tan A

  60. Given, cos4Acos2B+sin4Asin2B=1=sin2A+cos2A\frac{\cos^4 A}{\cos^2 B} + \frac{\sin^4 A}{\sin^2 B} = 1 = \sin^2A + \cos^2A

    cos4Acos2Bcos2A=sin2Asin4Asin2B=0\Rightarrow \frac{\cos^4 A}{\cos^2 B} - \cos^2A = \sin^2A - \frac{\sin^4A}{\sin^2B} = 0

    cos2A(cos2Acos2B)cos2B=sin2A(sin2Bsin2A)sin2B\Rightarrow \frac{\cos^2A(\cos^2A - \cos^2B)}{\cos^2B} = \frac{\sin^2A(\sin^2B - \sin^2A)}{\sin^2B}

    cos2A(cos2Acos2B)cos2B=sin2A(1cos2B1+cos2A)sin2B\Rightarrow \frac{\cos^2A(\cos^2A - \cos^2B)}{\cos^2B} = \frac{\sin^2A(1 - \cos^2B - 1 + \cos^2A)}{\sin^2B}

    (cos2Acos2B)(cos2Acos2Bsin2Asin2B)=0\Rightarrow (\cos^2A - \cos^2B )\left(\frac{\cos^2A}{\cos^2B} - \frac{\sin^2A}{\sin^2B}\right) = 0

    When cos2Acos2B=0,cos2A=cos2B\cos^2A - \cos^2B = 0, \cos^2A = \cos^2B

    When cos2Acos2Bsin2Asin2B=0\frac{\cos^2A}{\cos^2B} - \frac{\sin^2A}{\sin^2B} = 0

    cos2Asin2B=sin2Acos2Bcos2A(1cos2B)=(1cos2A)cos2B\cos^2A\sin^2B = \sin^2A\cos^2B \Rightarrow \cos^2A(1 - \cos^2B) = (1 - \cos^2A)\cos^2B

    cos2A=cos2Bsin2A=sin2B\cos^2A = \cos^2B \Rightarrow \sin^2A = \sin^2B

    1. L.H.S. =sin4A+sin4B=(sin2Asin2B)2+2sin2Asin2B=2sin2Asin2B== \sin^4A + \sin^4B = (\sin^2A - \sin^2B)^2 + 2\sin^2A\sin^2B = 2\sin^2A\sin^2B = R.H.S.

    2. L.H.S. =cos4Bcos2A+sin4Bsin2A= \frac{\cos^4 B}{\cos^2 A} + \frac{\sin^4 B}{\sin^2 A}

      =cos4Bcos2B+sin4Bsin2B=cos2B+sin2B=1== \frac{\cos^4 B}{\cos^2 B} + \frac{\sin^4 B}{\sin^2 B} = \cos^2B + \sin^2B = 1 = R.H.S.

  61. Given, cosA+sinA=2cosA\cos A + \sin A = \sqrt{2}\cos A

    Squaring both sides

    1+2sinAcosA=2cos2A1 + 2\sin A\cos A = 2\cos^2A

    22cos2A=2sin2A=12sinAcosA=sin2A+cos2A2sinAcosA=(cosAsinA)22 - 2\cos^2A = 2\sin^2A = 1 - 2\sin A\cos A = \sin^2A + \cos^2A - 2\sin A\cos A = (\cos A - \sin A)^2

    cosAsinA=±2sinA\cos A - \sin A = \pm \sqrt{2}\sin A

  62. Given, acosAbsinA=c,a\cos A - b\sin A = c, and let asinA+bcosA=xa\sin A + b\cos A = x

    Squaring and adding, we get

    a2cos2A+b2sin2A2abcosAsinA+a2sin2A+b2cos2A+2absinAcosA=c2+x2\Rightarrow a^2\cos^2A + b^2\sin^2A - 2ab\cos A\sin A + a^2\sin^2A + b^2\cos^2A + 2ab\sin A\cos A = c^2 + x^2

    a2(cos2A+sin2A)+b2(sin2A+cos2A)=c2+x2\Rightarrow a^2(\cos^2A + \sin^2A) + b^2(\sin^2A + \cos^2A) = c^2 + x^2

    a2+b2=c2+x2x=±a2+b2c2\Rightarrow a^2 + b^2 = c^2 + x^2 \Rightarrow x = \pm \sqrt{a^2 + b^2 - c^2}

  63. Given, 1sinA=1+sinA1 - \sin A = 1 + \sin A

    Multiplying both sides by 1sinA,1 - \sin A, we get

    (1sinA)2=1sin2A=cos2A(1 - \sin A)^2 = 1 - \sin^2A = \cos^2A

    1sinA=±cosA1 - \sin A = \pm \cos A

    Similarly if we multiply with 1+sin1 + \sin we can prove that

    1+sinA=±cosA1 + \sin A = \pm \cos A

  64. Let us solve these one by one:

    1. Given, sin4A+sin2A=1sin4A=1sin2A=cos2\sin^4A + \sin^2A = 1 \Rightarrow \sin^4A = 1 - \sin^2A = \cos^2

      sin4Acos4A=sec2A\frac{\sin^4A}{\cos^4A} = sec^2A

      1tan4A=cos2A\frac{1}{\tan^4A} = cos^2A

      Also from, sin4A=cos2Atan2A=cosec2A1tan2A=sin2A\sin^4A = \cos^2A \Rightarrow \tan^2A = \cosec^2A \Rightarrow \frac{1}{\tan^2A} = \sin^2A

      1tan4A+1tan2A=sin2A+cos2A=1\Rightarrow \frac{1}{\tan^4A} + \frac{1}{\tan^2A} = \sin^2A + \cos^2A = 1

    2. Given, sin4A+sin2A=1sin4A=1sin2A=cos2A\sin^4A + \sin^2A = 1 \Rightarrow \sin^4A = 1 - \sin^2A = \cos^2A

      sin2Acos2A=1sin2Atan2A=cosec2A\Rightarrow \frac{\sin^2A}{\cos^2A} = \frac{1}{\sin^2A} \Rightarrow \tan^2A = \cosec^2A

      tan2A=1+cot2A\tan^2A = 1 + \cot^2A

      Multiplying both sides with tan2A,\tan^2A, we get

      tan4A=tan2A+1\tan^4A = \tan^2A + 1

      tan4Atan2A=1\tan^4A - \tan^2A = 1

  65. Given, cos2sin2A=tan2Bcos2Asin2Acos2A+sin2A=tan2B\cos^2 - \sin^2 A = \tan^2 B \Rightarrow \frac{\cos^2A - \sin^2A}{\cos^2A + \sin^2A} = \tan^2B

    Dividing both numerator and denominator of L.H.S. with cos2A,\cos^2A, we get

    1tan2A1+tan2A=tan2B\frac{1 - \tan^2A}{1 + \tan^2A} = \tan^2B

    1tan2Atan2Btan2Atan2B=01 - \tan^2A - \tan^2B - \tan^2A\tan^2B = 0

    1tan2B=tan2A(1+tan2B)1tan2B1+tan2B=tan2A1 - \tan^2B = \tan^2A(1 + \tan^2B) \Rightarrow \frac{1 - \tan^2B}{1 + \tan^2B} = \tan^2A

    cos2Bsin2B=tan2A2cos2B1=tan2A\cos^2B - \sin^2B = \tan^2A \Rightarrow 2\cos^2B - 1 = \tan^2A

  66. We will prove this by induction. Let sinA+cosecA=2,\sin A + \cosec A = 2, thus it is true for n=1n = 1

    Squaring both sides sin2A+cosec2A+2sinAcosecA=22sin2A+cosec2A=2\sin^2A + \cosec^2A + 2\sin A\cosec A = 2^2 \Rightarrow \sin^2A + \cosec^2A = 2

    Thus, it is true for n=2n = 2 as well. Let it be true for n=m1n = m-1 and n=mn = m

    sinmA+cosecmA=2\sin^mA + \cosec^mA = 2

    Multiplying both sides with sinA+cosecA,\sin A + \cosec A, we get

    sinm+1A+cosecm+1A+sinmAcosecA+cosecmAsinA=22\sin^{m + 1}A + \cosec^{m + 1}A + \sin^mA\cosec A + \cosec^mA\sin A = 2^2

    sinm+1A+cosecm+1A+sinm1A+cosecm1A=4\sin^{m + 1}A + \cosec^{m + 1}A + \sin^{m - 1}A + \cosec^{m - 1}A = 4

    sinm+1A+cosecm+1=42=2\sin^{m + 1}A + \cosec^{m + 1} = 4 - 2 = 2

    Thus, we have proven it by induction.

  67. L.H.S. =secA+tan3AcosecA=secA(1+tan3AcosecAsecA)= \sec A + \tan^3A\cosec A = \sec A\left(1 + \tan^3A\frac{\cosec A}{\sec A}\right)

    =secA(1+tan3Acota)=secA(1+tan2A)=sec3A=(1+1e2)32=(2e2)32=\sec A(1 + \tan^3A\cot a) = \sec A(1 + \tan^2A) = \sec^3A = (1 + 1 - e^2)\frac{3}{2} = (2 - e^2)\frac{3}{2}

  68. Cross multiplying given equations, we have

    secAbrqc=tanApcqr=1qapb\frac{\sec A}{br - qc} = \frac{\tan A}{pc - qr} = \frac{1}{qa - pb}

    sec2Atan2A=1,(brqcaqpb)2(pcqaaqpb)2=1\therefore \sec^2A - \tan^2A = 1, \left(\frac{br - qc}{aq - pb}\right)^2 - \left(\frac{pc - qa}{aq - pb}\right)^2 = 1

    (brqa)2(pcar)2=(qapb)2(br - qa)^2 - (pc - ar)^2 = (qa - pb)^2

  69. Given, cosecAsinA=m1sinAsinA=m\cosec A - \sin A = m \Rightarrow \frac{1}{\sin A } - \sin A = m

    1sin2AsinA=mcos2AsinA=m\Rightarrow \frac{1 - \sin^2A}{\sin A} = m \Rightarrow \frac{\cos^2A}{\sin A} = m

    Also given, secAcosA=n1cosAcosA=n\sec A - \cos A = n \Rightarrow \frac{1}{\cos A} - \cos A = n

    1cos2AcosA=nsin2AcosA=n\Rightarrow \frac{1 - \cos^2A}{\cos A} = n \Rightarrow \frac{\sin^2A}{\cos A} = n

    We have sinA=cos2Am,\sin A = \frac{\cos^2A}{m}, putting this in derived equation

    cos3A=m2n\cos^3A = m^2n

    sinA=(m2n)23m=(mn2)13\therefore \sin A = \frac{(m^2n)^\frac{2}{3}}{m} = (mn^2)\frac{1}{3}

    Thus, sin2A+cos2A=1\sin^2A + \cos^2A = 1 gives us (m2n)23+(mn2)23=1(m^2n)^\frac{2}{3} + (mn^2)^\frac{2}{3} = 1

  70. Given, sec2A=4xy(x+y)2\sec^2A = \frac{4xy}{(x + y)^2}

    sec2A14xy(x+y)21\because \sec^2A \geq 1 \therefore \frac{4xy}{(x + y)^2}\geq 1

    (x+y)24xy(xy)20\Rightarrow (x + y)^2 \leq 4xy \Rightarrow (x - y)^2 \leq 0

    But for real xx and y,(xy)20y, (x - y)^2\nless 0

    (xy)2=0xy\therefore (x - y)^2 = 0 \therefore x - y

    Also, x+y0x0,y0x + y \neq 0 \Rightarrow x\neq 0, y \neq 0

  71. Given, sinA=x+1x,\sin A = x + \frac{1}{x}, squaring we get

    sin2A=x2+1x2+22\sin^2A = x^2 + \frac{1}{x^2} + 2 \geq 2

    which is not possible since sinA1\sin A \leq 1

  72. Given, secAtanA=p\sec A - \tan A = p

    1sinA=pcosA\Rightarrow 1 - \sin A = p\cos A

    Squaring, we obtain

    1+sin2A2sinA=p2cos2A=p2(1sin2A)1 + \sin^2A - 2\sin A = p^2\cos^2A = p^2(1 - \sin^2A)

    (1+p)sin2A2sinA+1p2=0(1 + p^)\sin^2A - 2\sin A + 1 - p^2 = 0

    sinA=1±p21+p2\sin A = \frac{1 \pm p^2}{1 + p^2}

    Now tantan and secA\sec A can be easily found.

  73. secA+tanA=4p2+14p±sec2A1\sec A + \tan A = \frac{4p^2 + 1}{4p} \pm \sqrt{\sec^2A - 1}

    =4p±((4p2+1)216p21)= 4p \pm \sqrt{\left(\frac{(4p^2 + 1)^2}{16p^2} - 1\right)}

    =2p= 2p or 12p\frac{1}{2p}

  74. Given, sinAsinB=p,cosAcosB=q\frac{\sin A}{\sin B} = p, \frac{\cos A}{\cos B} = q

    Squaring, we get

    sin2Asin2B=p2,cos2Acos2B=q2\frac{\sin^2A}{\sin^2B} = p^2, \frac{\cos^2A}{\cos^2B}= q^2

    sin2Asin2Bsin2B=p21,cos2Bcos2Acos2B=1q2\frac{\sin^2A - \sin^2B}{\sin^2B} = p^2 - 1, \frac{\cos^2B - \cos^2A}{\cos^2B} = 1 - q^2

    sin2Asin2Bsin2B=p21,sin2Asin2Bcos2B=1q2\frac{\sin^2A - \sin^2B}{\sin^2B} = p^2 - 1, \frac{\sin^2A - \sin^2B}{\cos^2B} = 1 - q^2

    Dividing, we obtain

    tan2B=±1q2p21\tan^2B = \pm \sqrt{\frac{1 - q^2}{p^2 - 1}}

    Dividing original equations

    tanAtanB=±pq1q2p21\frac{\tan A}{\tan B} = \pm \frac{p}{q}\sqrt{\frac{1 - q^2}{p^2 - 1}}

  75. sinA=2sinB,sinA.cosBcosA.sinB=3\sin A = \sqrt{2}\sin B, \frac{\sin A.\cos B}{\cos A. \sin B} = \sqrt{3}

    Substituting for sinA\sin A

    2sinBcosB12sin2AsinB=3\frac{\sqrt{2}\sin B\cos B}{\sqrt{1 - 2\sin^2A}\sin B} = \sqrt{3}

    Squaring, we get

    2cos2A=3(12sin2A)4sin2A1=02\cos^2 A = 3(1 - 2\sin^2A) \Rightarrow 4\sin^2A - 1 = 0

    sinA=±12,A=±45\Rightarrow \sin A = \pm \frac{1}{2}, \Rightarrow A = \pm 45^\circ

    Thus, B=±30B = \pm 30^\circ

  76. Given, tanA+cotA=2,1+tan2A=2tanA(1tanA)2=0\tan A + \cot A = 2, 1 + \tan^2A = 2\tan A \Rightarrow (1 - \tan A)^2 = 0

    tanA=1sinA=±12\tan A = 1\Rightarrow \sin A = \pm\frac{1}{\sqrt{2}}

  77. m2n2=(tanA+sinA)2(tanAsinA)2=4tanAsinAm^2 - n^2 = (\tan A +\sin A)^2 - (\tan A - \sin A)^2 = 4\tan A\sin A

    4mn=4tan2Asin2A=4sin2Acos2Asin2A4\sqrt{mn} = 4\sqrt{\tan^2A - \sin^2A} = 4\sqrt{\frac{\sin^2A}{\cos^2A} - \sin^2A}

    =4sin2A(1cos2A)cos2A=4sin4Acos2A= 4\sqrt{\frac{\sin^2A(1 - \cos^2A)}{\cos^2A}} = 4\sqrt{\frac{\sin^4A}{\cos^2A}}

    =4tanAsinA= 4\tan A\sin A

  78. m21=2sinAcosAn(m21)=2sinAcosA(secA+cosecA)m^2 - 1 = 2\sin A\cos A \Rightarrow n(m^2 - 1) = 2\sin A\cos A(\sec A + \cosec A)

    =2sinA+2cosA=2m= 2\sin A + 2\cos A = 2m

  79. Given, xsin3A+ycos3A=sinAcosAx\sin^3 A + y\cos^3 A = \sin A\cos A

    (xsinA)sin2A+(ycosA)cos2A=sinAcosA(x\sin A)\sin^2A + (y\cos A)\cos^2A = \sin A\cos A

    Frpm other equation xsinA=ycosA,x\sin A = y\cos A, substituting this in above equation

    (xsinA)sin2A+(xsinA)cos2a=sinAcosA(x\sin A)\sin^2A + (x\sin A)\cos^2a = \sin A\cos A

    xsinA=sinAcosAx\sin A = \sin A\cos A

    x=cosAy=sinAx = \cos A \therefore y = \sin A

    Thus, x2+y2=1x^2 + y^2 = 1

  80. Given, sin2A=(x+y)24xy\sin^2A = \frac{(x + y)^2}{4xy}

    sin2A1(x+y)24xy(xy)20\sin^2A \leq 1 \Rightarrow (x + y)^2 \leq 4xy \Rightarrow (x - y)^2 \leq 0

    But for real xx and yy (xy)20(x - y)^2\nless 0

    x=y\Rightarrow x = y

    Also, xy0,x,y0xy \neq 0, x, y\neq 0