Sunil Raikwar
Last Activity: 10 Years ago
Hello student, first term of your question should be 1/cos1cos2 in place of 1/cos1,
check the solution of your question given below:
x=(1/cos1°cos2) +(1/cos2°cos3°) +(1/cos3°cos4°) +....+(1/cos44°cos45°)
sin1x=(sin1/cos2cos1°) +(sin1/cos2°cos3°) +(sin1/cos3°cos4°) +....+(sin1/cos44°cos45°)
sin1x=(sin(2-1)/cos0cos1°) +(sin(3-2)/cos2°cos3°) +(sin(4-3)/cos3°cos4°) +....+(sin(45-44)/cos44°cos45°)
sin1x=tan2-tan1+tan3-tan2+tan4-tan3..........................+tan45-tan44
sin1x=tan45-tan1=1-tan1