prove that

(1+cot theta + tan theta) (sin theta-cos theta) = sec theta /cosec square theta - cosec theta/sec square theta.

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Question

prove that
(1+cot theta + tan theta) (sin theta-cos theta) = sec theta
/cosec square theta - cosec theta/sec square theta

Manbar Singh · Accepted Answer

LHS = 1 + cot θ + tan θ sin θ - cos θ=1 + cos
θsin θ + sin θcos θ sin θ - cos θ=sin
θ cos θ + cos2θ + sin2θsin θ cos θ
sin θ - cos θ=sin θ cos θ + 1 sin θ - cos
θsin θ cos θ RHS = sec θcosec2θ - cosec
θsec2θ=1/cosθ1/sin2θ - 1/sin
θ1/cos2θ=sin2θcosθ - cos2θsin
θ=sin3θ - cos3θsin θ cos θ=sin θ -
cos θ sin2θ +cos2θ +sin θ cos θ sin θ
cos θ =sin θ - cos θ 1 + sin θ cos θ sin
θ cos θ so, LHS = RHS

prove that (1+cot theta + tan theta) (sin theta-cos theta) = sec theta /cosec square theta - cosec theta/sec square theta.

prove that

(1+cot theta + tan theta) (sin theta-cos theta) = sec theta /cosec square theta - cosec theta/sec square theta.