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x=(1/cos1°) + (1/cos2°cos3°) +(1/cos3°cos4°) +....+(1/cos44°cos45°) , then x(sin1) is ?

x=(1/cos1°) + (1/cos2°cos3°) +(1/cos3°cos4°) +....+(1/cos44°cos45°) ,
then x(sin1) is ?

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1 Answers

Sunil Raikwar
askIITians Faculty 45 Points
9 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

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