Asked by maila molina
HOW TO PROVE TANGENT THETA +(SECANT THETA--1 DIVIDE TANGENT THETA-SECANT THETA+1=TANGENT THETA+SECANT THETA
Answers
Answered by
Reiny
tanØ + secØ - 1 / tanØ - secØ + 1 = tanØ + secØ
Surely , you must have meant
(tanØ + secØ - 1) / (tanØ - secØ + 1) = tanØ + secØ
I will use x instead of Ø for easier typing
LS = (sinx/cosx + 1/cosx - cosx/cosx) / (sinx/cosx - 1/cosx + cosx/cosx)
= (sinx + 1 - cosx)/(sinx - 1 + cosx)
= (sinx + 1 - cosx)/(sinx - 1 + cosx) * (sinx - 1 - cosx)/(sinx - 1 - cosx)
carefully expanding this, and replacing 1 with sin^2 x + cos^2 x , I got
= -2sinx cosx/(2sin^2 x - 2sinx)
= cosx/(1 - sinx)
RS = sinx/cosx + 1/cosx
= (sinx + 1)/cosx
= (sinx + 1)/cosx * (sinx - 1)/(sinx - 1)
= (sin^2 x - 1)/(sinxcosx - cosx)
= -cos^2 x/(cosx(sinx - 1))
= cosx/(1 - sinx)
= LS
Whewww!!!!
There has to be an easier way.
Surely , you must have meant
(tanØ + secØ - 1) / (tanØ - secØ + 1) = tanØ + secØ
I will use x instead of Ø for easier typing
LS = (sinx/cosx + 1/cosx - cosx/cosx) / (sinx/cosx - 1/cosx + cosx/cosx)
= (sinx + 1 - cosx)/(sinx - 1 + cosx)
= (sinx + 1 - cosx)/(sinx - 1 + cosx) * (sinx - 1 - cosx)/(sinx - 1 - cosx)
carefully expanding this, and replacing 1 with sin^2 x + cos^2 x , I got
= -2sinx cosx/(2sin^2 x - 2sinx)
= cosx/(1 - sinx)
RS = sinx/cosx + 1/cosx
= (sinx + 1)/cosx
= (sinx + 1)/cosx * (sinx - 1)/(sinx - 1)
= (sin^2 x - 1)/(sinxcosx - cosx)
= -cos^2 x/(cosx(sinx - 1))
= cosx/(1 - sinx)
= LS
Whewww!!!!
There has to be an easier way.
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