Question

Prove: 1+cos-sin2/sin(1+cos) = cot

Answer 1

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

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

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

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:

$\sf \implies cot \theta$

$\therefore$

$\bold{LHS = RHS}$

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

Answer 2

Solution :-

$\longrightarrow \frac{1 + \cos - { \sin}^{2}}{ \sin(1 + \cos)} = \cot \\ \\\longrightarrow \frac{1 + \cos - {(1 - \cos)}^{2} }{ \sin(1 + \cos)} = \cot \\ \\\longrightarrow \frac{1 + \cos - (1 + \cos)(1 - \cos)}{ \sin(1 + \cos)} = \cot \\ \\\longrightarrow\frac{1 - (1 - \cos)}{ \sin} = \cot \\\\\longrightarrow\frac{1 -1 + \cos}{ \sin} = \cot \\ \\\longrightarrow \frac{ \cos}{ \sin} = \cot \\ \\ \longrightarrow\cot = \cot$

LHS = RHS

Hence Proved