Question

Prove: 1+cos-sin2/sin(1+cos) = cot​

Answer 1

➥To Proof:

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

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

➥LHS:

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

➥RHS:

$\sf \implies cot \theta$

$\therefore$

$\bold{LHS = RHS}$

Hence Proved

Answer 2

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

$\huge\color{yellow}{\underline{\underline{answer\::}}}$

Refers to the attachment....