Question

prove that 1+cos theta -sin2 theta / sin theta(1+cos theta)=cot theta

Answer 1

Answer:

$\bold{\underline{To \ Proof:}}$

$\sf \dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} = cot \theta$

$\bold{\underline{Proof:}}$

LHS:

$\sf \implies \dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} \\ \\ \sf {sin}^{2} \theta = 1 - {cos}^{2} \theta : \\ \sf \implies \dfrac{1 + cos \theta - (1 - {cos}^{2} \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf (1 - cos ^{2} \theta ) = ( {1}^{2} - cos ^{2} \theta ) = (1 - cos \theta )(1 + cos \theta ) : \\ \sf \implies \dfrac{1 + cos \theta - (1 - cos \theta )(1 + cos \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf \implies \dfrac{ \cancel{1 + cos \theta}(1 - (1 - cos \theta ))}{sin \theta( \cancel{1 + cos \theta})} \\ \\ \sf \implies \dfrac{1 - (1 - cos \theta )}{sin \theta} \\ \\ \sf \implies \dfrac{1 - 1 + cos \theta )}{sin \theta} \\ \\ \sf \implies \dfrac{cos \theta }{sin \theta} \\ \\ \sf \implies cot \theta$

RHS:

$\sf \implies cot \theta$

$\therefore$

$\bold{LHS = RHS}$

$\underline{ \sf Hence \: Proved}$