5. Trigonometrical Ratios Solutions#

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

  1. L.H.S. =1−cos⁡A1+cos⁡A= \sqrt{\frac{1- \cos A}{1 + \cos A}}

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

    =(1−cos⁡A)21−cos⁡2A= \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}}

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

    =1−cos⁡Asin⁡A=cosec⁡A−cot⁡A== \frac{1 - \cos A}{\sin A} = \cosec A - \cot A = R.H.S.

  2. L.H.S. 1+tan⁡2A+1+cot⁡2A\sqrt{1 + \tan^2A + 1 + \cot^2A} [∵sec⁡2A=1+tan⁡2A\because \sec^2A = 1 + \tan^2A and cosec⁡2A=1+cot⁡2A\cosec^2A = 1 + \cot^2 A ]

    =tan⁡2A+2tan⁡Acot⁡A+cot⁡2A= \sqrt{\tan^2A + 2\tan A\cot A + \cot^2A} [tan⁡Acot⁡A=1\tan A\cot A = 1 ]

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

  3. L.H.S =(1−sin⁡2Asin⁡A)(1−cos⁡2Acos⁡A)(sin⁡2A+cos⁡2Asin⁡Acos⁡A)= \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)

    =cos⁡2Asin⁡Asin⁡2Acos⁡A1sin⁡Acos⁡A=1= \frac{\cos^2A}{\sin A}\frac{\sin^2 A}{\cos A}\frac{1}{\sin A \cos A} = 1

  4. L.H.S. =cos⁡4A−sin⁡4A+1=(cos⁡2A+sin⁡2A)(cos⁡2A−sin⁡2A)+1= \cos^4A - \sin^4A + 1= (\cos^2A + \sin^2A)(\cos^2A - \sin^2A) + 1

    =cos⁡2A−sin⁡2A+1=2cos⁡2A= \cos^2A - \sin^2A + 1 = 2\cos^2A

  5. L.H.S. =(sin⁡A+cos⁡A)(1−sin⁡Acos⁡A)=(sin⁡A+cos⁡A)(sin⁡2A+cos⁡2A−sin⁡Acos⁡A)= (\sin A + \cos A)(1 - \sin A\cos A) = (\sin A + \cos A)(\sin^2A + \cos^2A - \sin A\cos A)

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

  6. L.H.S =sin⁡A1+cos⁡A+1+cos⁡Asin⁡A=sin⁡2A+(1+cos⁡A)2sin⁡A(1+cos⁡A)= \frac{\sin A}{1 + \cos A}+\frac{1 + \cos A}{\sin A} = \frac{\sin^2A + (1 + \cos A)^2}{\sin A(1 + \cos A)}

    =sin⁡2A+1+2cos⁡A+cos⁡2Asin⁡A(1+cos⁡A)=1+1+2cos⁡A(sin⁡A)(1+cos⁡A))= \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))}

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

  7. L.H.S. =sin⁡6A−cos6A=(sin⁡2A+cos⁡2A)3−3cos⁡4Asin⁡2A−3cos⁡2Asin⁡4A= \sin^6A - cos^6A = (\sin^2A + \cos^2A)^3 - 3\cos^4A\sin^2A - 3\cos^2A\sin^4A

    =1−3sin⁡2Acos⁡2A(cos⁡2A+sin⁡2A)=1−3sin⁡2Acos⁡2A== 1 - 3\sin^2A\cos^2A(\cos^2A + \sin^2A) = 1 - 3\sin^2A\cos^2A = R.H.S.

  8. L.H.S. =1−sin⁡A1+sin⁡A= \sqrt{\frac{1 - \sin A}{1 + \sin A}}

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

    =(1−sin⁡A)21−sin⁡2A=(1−sin⁡A)2cos⁡2A=(1−sin⁡Acos⁡A)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}

    =1−sin⁡Acos⁡A=sec⁡A−tan⁡A= \frac{1 - \sin A}{\cos A} = \sec A - \tan A

  9. L.H.S. =cosec⁡Acosec⁡A−1+cosec⁡Acosec⁡A+1=1sin⁡A1sin⁡A−1+1sin⁡A1sin⁡A+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}

    =11−sin⁡A+11+sin⁡A=1(1−sin⁡A)(1+sin⁡A)=1cos⁡2A=sec⁡2A== \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. =cosec⁡Atan⁡A+cot⁡A=1sin⁡Asin⁡Acos⁡A+cos⁡Asin⁡A= \frac{\cosec A}{\tan A + \cot A} = \frac{\frac{1}{\sin A}}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}

    =1sin⁡Asin⁡2A+cos⁡2Asin⁡Acos⁡A=1sin⁡A1sin⁡Acos⁡A=cos⁡A== \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. =(sec⁡A+cos⁡A)(sec⁡A−cos⁡A)=sec⁡2A−cos⁡2A=1+tan⁡2A−1+sin⁡2A=tan⁡2A+sin⁡2A== (\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. =1tan⁡A+cot⁡A=1sin⁡Acos⁡A+cos⁡Asin⁡A=1sin⁡2A+cos⁡2Asin⁡Acos⁡A=sin⁡Acos⁡A== \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 =1−tan⁡A1+tan⁡A=1−1cot⁡A1+1cot⁡A=cot⁡A−1cot⁡A+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+tan⁡2A1+cot⁡2A=1+sin⁡2Acos⁡2A1+cos⁡2Asin⁡A= \frac{1 + \tan^2A}{1 + \cot^2A} = \frac{1 + \frac{\sin^2A}{\cos^2A}}{1 + \frac{\cos^2A}{\sin^A}}

    =cos⁡2A+sin⁡2Acos⁡2Asin⁡2A+cos⁡2Asin⁡2A=sin⁡2Acos⁡2A== \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. =sec⁡A−tan⁡Asec⁡A+tan⁡A= \frac{\sec A - \tan A}{\sec A + \tan A}

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

    (sec⁡A−tan⁡A)2sec⁡2A−tan⁡2A=sec⁡2A−2tan⁡Asec⁡A+tan⁡A=1+tan⁡2A−2tan⁡Asec⁡A+tan⁡2A\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

    =1−2tan⁡Asec⁡A+2tan⁡2A== 1 - 2\tan A\sec A + 2\tan^2A = R.H.S.

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

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

    =sec⁡A+tan⁡Asec⁡2A=tan⁡2A=sec⁡A+tan⁡A== \frac{\sec A + \tan A}{\sec^2A = \tan^2A} = \sec A + \tan A = R.H.S.

  17. L.H.S. =tan⁡A1−cot⁡A+cot⁡A1−tan⁡A= \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}

    =sin⁡Acos⁡A1−cos⁡Asin⁡A+cos⁡Asin⁡A1−sin⁡Acos⁡A= \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}

    =sin⁡2Acos⁡A(sin⁡A−cos⁡A)+cos⁡2Asin⁡A(cos⁡A−sin⁡A)= \frac{\sin^2A}{\cos A(\sin A - \cos A)} + \frac{\cos^2A}{\sin A(\cos A - \sin A)}

    =sin⁡3A−cos⁡3Asin⁡Acos⁡A(sin⁡A−cos⁡A)=sin⁡2A+cos⁡2A+sin⁡Acos⁡Asin⁡Acos⁡A= \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+sin⁡Acos⁡Asin⁡Acos⁡A=1+cosec⁡Asec⁡A== \frac{1 + \sin A \cos A}{\sin A\cos A} = 1 + \cosec A\sec A = R.H.S.

  18. L.H.S. =cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=cos⁡A1−sin⁡Acos⁡A+sin⁡A1−cos⁡Asin⁡A= \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}}

    =cos⁡2Acos⁡A−sin⁡A+sin⁡2Asin⁡A−cos⁡A=cos⁡2A−sin⁡2Acos⁡A−sin⁡A= \frac{\cos^2A}{\cos A - \sin A} + \frac{\sin^2A}{\sin A - \cos A} = \frac{\cos^2A - \sin^2A}{\cos A - \sin A}

    =cos⁡A+sin⁡A== \cos A + \sin A = R.H.S.

  19. L.H.S. =(sin⁡A+cos⁡A)(tan⁡A+cot⁡A)=(sin⁡A+cos⁡A)(sin⁡Acos⁡A+cos⁡Asin⁡A)= (\sin A + \cos A)(\tan A + \cot A) = (\sin A + \cos A)(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})

    =(sin⁡A+cos⁡A)sin⁡2A+cos⁡2Asin⁡Acos⁡A=sin⁡A+cos⁡Asin⁡Acos⁡A=sec⁡A+cosec⁡A== (\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. =sec⁡4A−sec⁡2A=(1+tan⁡2A)−1−tan⁡2A=1+2tan⁡2A+tan⁡4A−1−tan⁡2A=tan⁡4A+tan⁡2A== \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. =cot⁡4A+cot⁡2A=(cosec⁡2A−1)1+cosec⁡2A−1=cosec⁡4A−2cosec⁡2A+1+cosec⁡2A−1=cosec⁡4A−cosec⁡2A== \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. =cosec⁡2A−1=cot⁡2A=cot⁡A=cos⁡Asin⁡A=cos⁡Acosec⁡A== \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. =sec⁡2Acosec⁡2A=(1+tan⁡2A)(1+cot⁡2A)=1+tan⁡2A+cot⁡2A+tan⁡2Acos⁡2A=2+tan⁡2A+cot⁡2A== \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. =tan⁡2A−sin⁡2A=sin⁡2Acos⁡2A−sin⁡2A=sin⁡2A−sin⁡2Acos⁡2Acos⁡2A=sin⁡2A(1−cos⁡2A)sec⁡2A=sin⁡4Asec⁡2A== \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+cot⁡A−cosec⁡A)(1+tan⁡A+sec⁡A)=(1+cos⁡Asin⁡A−1sin⁡A)(1+sin⁡Acos⁡A+1cos⁡A)= (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)

    =(sin⁡A+cos⁡A−1)(sin⁡A+cos⁡A+1)sin⁡Acos⁡A=(sin⁡A+cos⁡A)2−1sin⁡Acos⁡A= \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}

    =sin⁡2A+cos⁡2A+sin⁡Acos⁡A−1sin⁡Acos⁡A=1+2sin⁡Acos⁡A−1sin⁡Acos⁡A=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. =cot⁡Acos⁡Acot⁡A+cos⁡A=cos⁡2Asin⁡Acos⁡Asin⁡A+cos⁡A= \frac{\cot A\cos A}{\cot A + \cos A} = \frac{\frac{\cos^2 A}{\sin A}}{\frac{\cos A}{\sin A} + \cos A}

    =cos⁡2Asin⁡Asin⁡Acos⁡A(1+sin⁡A)=cos⁡A1+sin⁡A= \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 1−sin⁡Acos⁡A\frac{1 - \sin A}{\cos A}

    cos⁡A1+sin⁡A=cos⁡A(1−sin⁡A)1−sin⁡2A=1−sin⁡Acos⁡A=\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. =cot⁡A+tan⁡Bcot⁡B+tan⁡A=1tan⁡A+tan⁡B1tan⁡B+tan⁡A= \frac{\cot A + \tan B}{\cot B + \tan A} = \frac{\frac{1}{\tan A} + \tan B}{\frac{1}{\tan B} + \tan A}

    1+tan⁡Atan⁡Btan⁡A1+tan⁡Atan⁡Btan⁡B=tan⁡Btan⁡A=cot⁡Atan⁡B=\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. =(1sec⁡2A−cos⁡2A+1cosec⁡2A−sin⁡2A)cos⁡2Asin⁡2A= \left(\frac{1}{\sec^2 A - \cos^2A} + \frac{1}{\cosec^2A - \sin^2A}\right)\cos^2A\sin^2A

    =(11cos⁡2A−cos⁡2A+11sin⁡2A−sin⁡2A)cos⁡2Asin⁡2A= \left(\frac{1}{\frac{1}{\cos^2A} - \cos^2A} + \frac{1}{\frac{1}{\sin^2A} - \sin^2A}\right)\cos^2A\sin^2A

    =(cos⁡2A1−cos⁡4A+sin⁡2A1−sin⁡4)cos⁡2Asin⁡2A= \left(\frac{\cos^2A}{1 - \cos^4A} + \frac{\sin^2A}{1 - \sin^4}\right)\cos^2A\sin^2A

    =(cos⁡2A(1−cos⁡2A)(1+cos⁡2A)+sin⁡2A(1−sin⁡2A)(1+sin⁡2A))cos⁡2Asin⁡2A= \left(\frac{\cos^2A}{(1 - \cos^2A)(1 + \cos^2A)} + \frac{\sin^2A}{(1 - \sin^2A)(1 + \sin^2A)}\right)\cos^2A\sin^2A

    =(cos⁡2Asin⁡2A(1+cos⁡2A)+sin⁡2Acos⁡2A(1+sin⁡2A))cos⁡2Asin⁡2A= \left(\frac{\cos^2A}{\sin^2A(1 + \cos^2A)} + \frac{\sin^2A}{\cos^2A(1 + \sin^2A)}\right)\cos^2A\sin^2A

    =(cos⁡4A(1+sin⁡2A)+sin⁡4A(1+cos⁡2A)sin⁡2Acos⁡2A(1+cos⁡2A))(1+sin⁡2A))cos⁡2Asin⁡2A= \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

    =(cos⁡4A+sin⁡4A+cos⁡2Asin⁡2A(cos⁡2A+sin⁡2A)(1+cos⁡2A)(1+sin⁡2A))= \left(\frac{\cos^4A + \sin^4A + \cos^2A\sin^2A(\cos^2A + \sin^2A)}{(1 +\cos^2A)(1 + \sin^2A)}\right)

    =((cos⁡2A+sin⁡2A)2−cos⁡2Asin⁡2A1+sin⁡2A+cos⁡2A+sin⁡2Acos⁡2A)= \left(\frac{(\cos^2A + \sin^2A)^2 - \cos^2A\sin^2A}{1 + \sin^2A + \cos^2A + \sin^2A\cos^2A}\right)

    =1−sin⁡2Acos⁡2A2+sin⁡2Acos⁡2A== \frac{1 - \sin^2A\cos^2A}{2 + \sin^2A\cos^2A} = R.H.S.

  29. L.H.S. =sin⁡8A−cos⁡8A=(sin⁡4A−cos⁡4A)(sin⁡4A+cos⁡4A)= \sin^8A - \cos^8A = (\sin^4A - \cos^4A)(\sin^4A + \cos^4A)

    =(sin⁡2A+cos⁡2A)(sin⁡2A−cos⁡2A)((sin⁡2A+cos⁡2A)2−2sin⁡2Acos⁡2A)= (\sin^2A + \cos^2A)(\sin^2A - \cos^2A)((\sin^2A + \cos^2A)^2 - 2\sin^2A\cos^2A)

    =(sin⁡2A−cos⁡2A)(1−2sin⁡2Acos⁡2A)== (\sin^2A - \cos^2A)(1 - 2\sin^2A\cos^2A) = R.H.S.

  30. L.H.S. =cos⁡Acosec⁡A−sin⁡Asec⁡Acos⁡A+sin⁡A=cos⁡Asin⁡A−sin⁡Acos⁡Acos⁡A+sin⁡A= \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}

    =cos⁡2A−sin⁡2Asin⁡Acos⁡A(cos⁡A+sin⁡A)=cos⁡Asin⁡Asin⁡Acos⁡A=cosec⁡A−sec⁡A== \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, 1cosec⁡A−cot⁡A−1sin⁡A=1sin⁡A−1cosec⁡A+cot⁡A\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 1cosec⁡A−cot⁡A+1cosec⁡A+cot⁡A=2sin⁡A\frac{1}{\cosec A - \cot A} + \frac{1}{\cosec A + \cot A} = \frac{2}{\sin A}

    L.H.S. =11sin⁡A−cos⁡Asin⁡A+11sin⁡A+cos⁡Asin⁡A= \frac{1}{\frac{1}{\sin A} - \frac{\cos A}{\sin A}} + \frac{1}{\frac{1}{\sin A} + \frac{\cos A}{\sin A}}

    =sin⁡A1−cos⁡A−sin⁡A1+cos⁡A=sin⁡A(1+cos⁡A+1−cos⁡A)1−cos⁡2A=2sin⁡Asin⁡2A=2sin⁡A== \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. =tan⁡A+sec⁡A−1tan⁡A−sec⁡A+1=sin⁡A+1−cos⁡Acos⁡Asin⁡A+cos⁡A−1)cos⁡A= \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+sin⁡A−cos⁡Asin⁡A+cos⁡A−1=(1+sin⁡A−cos⁡A)(sin⁡A+cos⁡A−1)(sin⁡A+cos⁡A−1)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+sin⁡Acos⁡A== \frac{1 + \sin A}{\cos A} = R.H.S.

  33. L.H.S. =(tan⁡A+cosec⁡B)2−(cot⁡B−sec⁡A)2=tan⁡2A+cosec⁡2B+2tan⁡Acosec⁡B−cot⁡2B−sec⁡2A+2sec⁡Acot⁡B= (\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

    =(cosec⁡2B−cot⁡2B)−(sec2A−tan2A)+2tan⁡Acot⁡B(cosec⁡Bcot⁡B+sec⁡Atan⁡A)= (\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)

    =1−1+2tan⁡Acot⁡B(sec⁡B+cosec⁡A)== 1 - 1 + 2\tan A\cot B\left(\sec B + \cosec A\right) = R.H.S.

  34. L.H.S. =2sec⁡2A−sec⁡4A−2cosec⁡2A+cosec⁡4A=2(1+tan⁡2A)−(1+tan⁡2A)2−2(1+cot⁡2A)+(1+cot⁡2A)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+2tan⁡2A−1−2tan⁡2A+tan⁡4A−2−2cot⁡2A+1+2cot⁡2A+cot⁡4A=cot⁡4A−tan⁡4A== 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. =(sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2=sin⁡2A+cosec⁡2A+2sin⁡Acosec⁡A+cos⁡2A+sec⁡2A+2sec⁡Acosec⁡A= (\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

    =(sin⁡2A+cos⁡2A)+4+cosec⁡2A+sec⁡2A=1+4+1+cot⁡2A+1+tan⁡2A=cot⁡2A+tan⁡2A+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 =(cosec⁡A+cot⁡A)(1−sin⁡A)−(sec⁡A+tan⁡A)(1−cos⁡A)= (\cosec A + \cot A)(1 - \sin A) - (\sec A + \tan A)(1 - \cos A)

    =(1sin⁡A+cos⁡Asin⁡A)(1−sin⁡A)−(1cos⁡A+sin⁡Acos⁡A)(1−cos⁡A)= \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+cos⁡A)(1−sin⁡A)sin⁡A−(1+sin⁡A)(1−cos⁡A)cos⁡A= \frac{(1 + \cos A)(1 - \sin A)}{\sin A} - \frac{(1 + \sin A)(1 - \cos A)}{\cos A}

    =cos⁡A(1+cos⁡A−sin⁡A−cos⁡Asin⁡A)−sin⁡A(1+sin⁡A−cos⁡A−sin⁡Acos⁡A)sin⁡Acos⁡A= \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}

    =(1−sin⁡Acos⁡A)(cos⁡A−sin⁡A)+(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)sin⁡Acos⁡A= \frac{(1 - \sin A\cos A)(\cos A - \sin A) + (\cos A - \sin A)(\cos A + \sin A)}{\sin A\cos A}

    =(cos⁡A−sin⁡A)(1−cos⁡Asin⁡A+cos⁡A+sin⁡A)sin⁡Acos⁡A= \frac{(\cos A - \sin A)(1 - \cos A\sin A + \cos A + \sin A)}{\sin A\cos A}

    =((cos⁡A−sin⁡A)sin⁡Acos⁡A)(2−1−cos⁡Asin⁡A+cos⁡A+sin⁡A)= \left(\frac{(\cos A - \sin A)}{\sin A\cos A}\right)(2 - 1 - \cos A\sin A + \cos A + \sin A)

    =(cosec⁡A−sec⁡A)(2−(1−cos⁡A)(1−sin⁡A))== (\cosec A - \sec A)(2 - (1 - \cos A)(1 - \sin A)) = R.H.S.

  37. L.H.S =(1+cot⁡A+tan⁡A)(sin⁡A−cos⁡A)=(1+cosec⁡Asec⁡A+sec⁡Acosec⁡A)(1cosec⁡A−1sec⁡A)= (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)

    =(sec⁡Acosec⁡A+sec⁡A∣cosec⁡Asec⁡Acosec⁡A)(sec⁡A−cosec⁡Asec⁡Acosec⁡a)= \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)

    =sec⁡3A−cosec⁡3Asec⁡2Acosec⁡2A=sec⁡Acosec⁡2A−cosec⁡Asec⁡2A== \frac{\sec^3A - \cosec^3A}{\sec^2A\cosec^2A} = \frac{\sec A}{\cosec^2A} - \frac{\cosec A}{\sec^2A} = R.H.S.

  38. Given, 1sec⁡A−tan⁡A−1cos⁡A=1cos⁡A−1sec⁡A+tan⁡A\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 1sec⁡A−tan⁡A+1sec⁡A+tan⁡A=2cos⁡A\frac{1}{\sec A - \tan A} + \frac{1}{\sec A + \tan A} = \frac{2}{\cos A}

    L.H.S. =sec⁡A+tan⁡A+sec⁡A−tan⁡Asec⁡2A−tan2A=2sec⁡A=2cos⁡A== \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(sin⁡A−cos⁡A)4+4(sin⁡6A+cos⁡6A)+6(sin⁡A+cos⁡A)2= 3(\sin A - \cos A)^4 + 4(\sin^6 A + \cos^6 A) + 6(\sin A + \cos A)^2

    =3[(sin⁡A−cos⁡A)2]2+4[(sin⁡2A)3+(cos⁡2A)3]+6(sin⁡2A+cos⁡2A+2cos⁡Asin⁡A)= 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[(sin⁡2A+cos⁡2A−2sin⁡Acos⁡A)]2+4(sin⁡2A+cos⁡2A)(sin⁡4A+cos⁡4A−sin⁡2Acos⁡2A)+6(1+2cos⁡Asin⁡A)= 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(1−2sin⁡Acos⁡A)2+4(sin⁡4A+cos⁡4A−sin⁡2Acos⁡2A)+6(1+2cos⁡Asin⁡A)= 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+4sin⁡2Acos⁡2A−4cos⁡Asin⁡A)+4[(sin⁡2A+cos⁡2A)2−2sin⁡2Acos⁡2A−sin⁡2Acos⁡2A]+6(1+2sin⁡Acos⁡A)= 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+12sin⁡2Acos⁡2A−12cos⁡Asin⁡A+4(1−3sin⁡2Acos⁡2A)+6+12sin⁡Acos⁡A= 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+cos⁡A1−cos⁡A= \sqrt{\frac{1 + \cos A}{1 - \cos A}}

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

    =(1+cos⁡A)21−cos⁡2A=(1+cos⁡Asin⁡A)2=cosec⁡A+cot⁡A== \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. =cos⁡A1+sin⁡A+cos⁡A1−sin⁡A= \frac{\cos A}{1 + \sin A} + \frac{\cos A}{1 - \sin A}

    cos⁡A(1−sin⁡A)+cos⁡A(1+sin⁡A)1−sin⁡2A\frac{\cos A(1 - \sin A) + \cos A(1 + \sin A)}{1 - \sin^2A}

    =2cos⁡Acos⁡2A=2sec⁡A== \frac{2\cos A}{\cos^2A} = 2\sec A = R.H.S.

  42. L.H.S. =tan⁡Asec⁡A−1+tan⁡Asec⁡A+1= \frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1}

    =tan⁡Asec⁡A+tan⁡A+tan⁡Asec⁡A−tan⁡A(sec⁡2A−1)= \frac{\tan A\sec A + \tan A + \tan A\sec A - \tan A}{(\sec^2A - 1)}

    =2tan⁡Asec⁡Atan⁡2A=2cosec⁡A== \frac{2\tan A\sec A}{\tan^2 A} = 2\cosec A = R.H.S.

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

    =1+sin⁡A−1+sin⁡A1−sin⁡2A=2sin⁡Acos⁡2A=2sec⁡Atan⁡A== \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+tan⁡2A1+cot⁡2A=1+sin⁡2Acos⁡2A1+cos⁡2Asin⁡2A= \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{1 + \frac{\sin^2A}{\cos^2A}}{1 + \frac{\cos^2A}{\sin^2A}}

    =sin⁡2A+cos⁡2Acos⁡2Asin⁡2A+cos⁡2Asin⁡2A=sin⁡2Acos⁡2A=tan⁡2A= \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. =(1−tan⁡A1−cot⁡A)2=(1−sin⁡Acos⁡A1−cos⁡Asin⁡A)= \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)

    =(cos⁡A−sinAcos⁡Asin⁡A−cos⁡Asin⁡A)2= \left(\frac{\frac{\cos A - sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}}\right)^2

    =sin⁡2Acos⁡2A=tan⁡2A= \frac{\sin^2A}{\cos^2A} = \tan^2A

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

  45. L.H.S. =1+2tan⁡2Acos⁡2A=1+2sin⁡2Acos⁡4A=cos⁡4A+2sin⁡2Acos⁡4A= 1 + \frac{2\tan^2 A}{\cos^2 A} = 1 + \frac{2\sin^2A}{\cos^4A} = \frac{\cos^4A + 2\sin^2A}{\cos^4A}

    =(1−sin⁡2A)2+2sin⁡2Acos⁡4A=1−2sin⁡2A+sin⁡4A+2sin⁡2Acos⁡4A= \frac{(1 - \sin^2A)^2 + 2\sin^2A}{\cos^4A} = \frac{1 - 2\sin^2A + \sin^4A + 2\sin^2A}{\cos^4A}

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

  46. L.H.S. =(1−sin⁡A−cos⁡A)2=1−2cos⁡A−2sin⁡A+2sin⁡Acos⁡A+sin⁡2A+cos⁡2A= (1 - \sin A - \cos A)^2 = 1 - 2\cos A - 2\sin A + 2\sin A\cos A + \sin^2A + \cos^2A

    =2−2cos⁡A−2sin⁡A+2sin⁡Acos⁡A=2(1−cos⁡A)(1−sin⁡A)== 2 - 2\cos A - 2\sin A + 2\sin A\cos A = 2(1 - \cos A)(1 - \sin A) = R.H.S.

        1. =cot⁡A+cosec⁡A−1cot⁡A−cosec⁡A+1= \frac{\cot A + \cosec A - 1}{\cot A - \cosec A + 1}

    =cos⁡A+1−sin⁡Acos⁡A+sin⁡A−1= \frac{\cos A + 1 - \sin A}{\cos A + \sin A - 1}

    Multiplying and dividing with cos⁡A−(1−sin⁡A),\cos A - (1 - \sin A), we get

    =cos⁡2A−1+2sin⁡A−sin⁡2Acos⁡2A+sin⁡2A+1−2cos⁡A−2sin⁡A+2sin⁡Acos⁡A= \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}

    =2sin⁡A−2sin⁡2A2(1−cos⁡A)(1−sin⁡A)=sin⁡A1−cos⁡A= \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+cos⁡A,1 + \cos A, we get

    =sin⁡A(1+cos⁡A)1−cos⁡2A=1+cos⁡Asin⁡A== \frac{\sin A(1 + \cos A)}{1 - \cos^2A} = \frac{1 + \cos A}{\sin A} = R.H.S.

  48. L.H.S. =(sin⁡A+sec⁡A)2+(cos⁡A+cosec⁡A)2= (\sin A + \sec A)^2 + (\cos A + \cosec A)^2

    =(sin⁡Acos⁡A+1cos⁡A)2+(cos⁡Asin⁡A+1sin⁡A)2= \left(\frac{\sin A\cos A + 1}{\cos A}\right)^2 + \left(\frac{\cos A\sin A + 1}{\sin A}\right)^2

    =(1+sin⁡Acos⁡A)2(1cos⁡2A+1sin⁡2A)= (1 + \sin A\cos A)^2 \left(\frac{1}{\cos^2A} + \frac{1}{\sin^2A}\right)

    =(1+sin⁡Acos⁡A)2(sin⁡2A+cos⁡2Asin⁡2Acos⁡2A)= (1 + \sin A\cos A)^2\left(\frac{\sin^2A + \cos^2A}{\sin^2A\cos^2A}\right)

    =(1+sin⁡Acos⁡Asin⁡Acos⁡A)2=(1+sec⁡Acosec⁡A)2== \left(\frac{1 + \sin A\cos A}{\sin A\cos A}\right)^2 = (1 + \sec A\cosec A)^2 = R.H.S.

  49. L.H.S. =2sin⁡Atan⁡A(1−tan⁡A)+2sin⁡Asec⁡2A(1+tan⁡A)2= \frac{2\sin A\tan A(1 - \tan A) + 2\sin A\sec^2A}{(1 + \tan A)^2}

    =2sin⁡A(tan⁡A−tan⁡2A+sec⁡2A)(1+tan⁡A)2=2sin⁡A(1+tan⁡A)(1+tan⁡A)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}

    =2sin⁡A1+tan⁡A== \frac{2\sin A}{1 + \tan A} = R.H.S.

  50. Given, 2sin⁡A=2−cos⁡A2\sin A = 2 - \cos A

    cos⁡A=2−2sin⁡A,\cos A = 2 - 2\sin A, squaring both sides we, get cos⁡2A=4−8sin⁡A+4sin⁡2A\cos^2A = 4 - 8\sin A + 4\sin^2A

    1−sin⁡2A=4−8sin⁡A+4sin⁡2A⇒5sin⁡2A−8sin⁡A+3=01 - \sin^2A = 4 - 8\sin A + 4\sin^2A \Rightarrow 5\sin^2A - 8\sin A + 3 = 0

    5sin⁡2A−5sin⁡A−3sin⁡A+3=0⇒sin⁡A=1,355\sin^2A - 5\sin A - 3\sin A + 3 = 0 \Rightarrow \sin A = 1, \frac{3}{5}

  51. Given 8sin⁡A=4+cos⁡A⇒8sin⁡A−4=cos⁡A8\sin A = 4 + \cos A \Rightarrow 8\sin A - 4 = \cos A

    Squaring both sides, we get

    64sin⁡2A−64sin⁡A+16=cos⁡2A=1−sin⁡2A64\sin^2A - 64\sin A + 16 = \cos^2A = 1 - \sin^2A

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

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

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

  52. Given, tan⁡A+sec⁡A=1.5\tan A + \sec A = 1.5

    ⇒1+sin⁡A=1.5cos⁡A⇒2+2sin⁡A=3cos⁡A\Rightarrow 1 + \sin A = 1.5 \cos A \Rightarrow 2 + 2\sin A = 3\cos A

    Squaring both sides, we get

    4+8sin⁡A+4sin⁡2A=9−9sin⁡2A4 + 8\sin A + 4\sin^2A = 9 - 9\sin^2A

    13sin⁡2A+8sin⁡A−5=013\sin^2A + 8\sin A - 5 =0

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

  53. Given, cot⁡A+cosec⁡A=5⇒1+cos⁡A=5sin⁡A\cot A + \cosec A = 5 \Rightarrow 1 + \cos A = 5\sin A

    Squarig both sides, we get

    1+2cos⁡A+cos⁡2A=25sin⁡2A=25(1−cos⁡2A)1 + 2\cos A + \cos^2A = 25\sin^2A = 25(1 - \cos^2A)

    26cos⁡2A+2cos⁡A−24=026\cos^2A + 2\cos A - 24 = 0

    26cos⁡2A+26cos⁡A−24cos⁡A−24=026\cos^2A + 26\cos A - 24\cos A - 24 = 0

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

  54. 3sec⁡4A+8=10sec⁡2A⇒3(sec2A)2+8=10(1+tan⁡2A)3\sec^4 A + 8 = 10\sec^2A \Rightarrow 3(sec^2A)^2 + 8 = 10(1 + \tan^2A)

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

    3+6tan⁡2A+3tan⁡4A+8=10+10tan⁡2A3 + 6\tan^2A + 3\tan^4A + 8 = 10 + 10\tan^2A

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

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

  55. Given, tan⁡2A+sec⁡A=5⇒sec⁡2A−1+sec⁡A=5\tan^2A + \sec A = 5 \Rightarrow \sec^2A - 1 + \sec A = 5

    sec⁡2A+sec⁡A−6=0\sec^2A + \sec A - 6 = 0

    sec⁡A=−3,2⇒cos⁡A=−13,12\sec A = -3, 2\Rightarrow \cos A = -\frac{1}{3}, \frac{1}{2}

  56. Given tan⁡A+cot⁡A=2⇒tan⁡A+1tan⁡A=2\tan A + \cot A = 2 \Rightarrow \tan A + \frac{1}{\tan A} = 2

    tan⁡2A−2tan⁡A+1=0\tan^2A - 2\tan A + 1 = 0

    tan⁡A=1⇒sin⁡A=12\tan A = 1 \Rightarrow \sin A = \frac{1}{\sqrt{2}}

  57. Given, sec⁡2A=2+2tan⁡A⇒tan⁡2A+1=2+2tan⁡A\sec^2A = 2 + 2\tan A \Rightarrow \tan^2A + 1 = 2 + 2\tan A

    tan⁡2A−2tan⁡A−1=0\tan^2A - 2\tan A - 1 = 0

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

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

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

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

  59. Given, 3sin⁡A+5cos⁡A=5,3\sin A + 5\cos A = 5, let 5sin⁡A−3cos⁡A=x5\sin A - 3\cos A = x

    Squaring and adding, we get

    9sin⁡2A+25cos⁡2A+30sin⁡Acos⁡A+25sin⁡2A+9cos⁡2A−30sin⁡Acos⁡A=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(sin⁡2A+cos⁡2A)+25(cos⁡2A+sin⁡2A)=25+x29(\sin^2A + \cos^2A) + 25(\cos^2A + \sin^2A) = 25 + x^2

    34=25+x2⇒x2=9⇒x=±334 = 25 + x^2 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3

  60. Given, sec⁡A+tan⁡A=sec⁡A−tan⁡A\sec A + \tan A = \sec A - \tan A

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

    (sec⁡A+tan⁡A)2=sec⁡2A−tan⁡2A=1(\sec A + \tan A)^2 = \sec^2A - \tan^2A = 1