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(-tan(theta) %2B sec(theta) %2B 1)/(tan(theta) %2B sec(theta) %2B 1)=(-sin(theta) %2B 1)/cos(theta) |
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Answer» Let theta = x LHS:secx - tanx + 1/secx + tanx + 1 = secx - tanx + (sec^2x - tan^2x)/ (secx + tanx + 1) = (secx - tanx) + (secx - tanx)(secx + tanx)/(secx + tanx + 1) = (secx - tanx)(secx + tanx + 1)/ (secx + tanx + 1) = secx - tanx = 1/cosx - sinx/cosx = (1 - sinx) /cosx = RHS Hence proved |
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