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\frac \tan \theta 1 - \cot \theta %2B \frac \cot \theta 1 - \tan \theta = \tan \theta %2B \cot \theta %2B 1 |
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Answer» Let theta = x LHS:tanx/1−cotx + cotx/1 - tanx = tan^2 x/(tanx - 1) + 1/tanx( 1 - tanx) = tan^3 x - 1/tanx(tanx - 1) = tan^2 x + tanx + 1/tanx = tanx + 1 + 1/tanx = tanx + cotx + = RHS Hence proved |
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