Question

prove it : 1+cos-sin2/sin(1+cos) = cot ​

Answer 1

To Proof:

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

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

LHS:

$\begin{gathered}\sf\red\implies \dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} \\ \\ \sf {sin}^{2} \theta = 1 - {cos}^{2} \theta : \\ \sf \pink\implies \dfrac{1 + cos \theta - (1 - {cos}^{2} \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf\green(1 - cos ^{2} \theta ) = ( {1}^{2} - cos ^{2} \theta ) = (1 - cos \theta )(1 + cos \theta ) : \\ \sf\purple \implies \dfrac{1 + cos \theta - (1 - cos \theta )(1 + cos \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf \implies \dfrac{ \cancel{1 + cos \theta}(1 - (1 - cos \theta ))}{sin \theta( \cancel{1 + cos \theta})} \\ \\ \sf\orange \implies \dfrac{1 - (1 - cos \theta )}{sin \theta} \\ \\ \sf\red \implies \dfrac{1 - 1 + cos \theta )}{sin \theta} \\ \\ \sf\blue \implies\dfrac{cos \theta }{sin \theta} \\ \\ \sf\pink \implies cot \theta\end{gathered}$

RHS:

$\sf\green\implies cot \theta$

$\therefore$

$\bold\red{LHS = RHS}$

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

Answer 2

To Proof:

$\sf\dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} </p><p>sinθ(1+cosθ)$

1+cosθ−sin 2 θ

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

LHS:

$\begin{gathered}\begin{gathered}\sf\red\implies \dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} \\ \\ \sf {sin}^{2} \theta = 1 - {cos}^{2} \theta : \\ \sf \pink\implies \dfrac{1 + cos \theta - (1 - {cos}^{2} \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf\green(1 - cos ^{2} \theta ) = ( {1}^{2} - cos ^{2} \theta ) = (1 - cos \theta )(1 + cos \theta ) : \\ \sf\purple \implies \dfrac{1 + cos \theta - (1 - cos \theta )(1 + cos \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf \implies \dfrac{ \cancel{1 + cos \theta}(1 - (1 - cos \theta ))}{sin \theta( \cancel{1 + cos \theta})} \\ \\ \sf\orange \implies \dfrac{1 - (1 - cos \theta )}{sin \theta} \\ \\ \sf\red \implies \dfrac{1 - 1 + cos \theta )}{sin \theta} \\ \\ \sf\blue \implies\dfrac{cos \theta }{sin \theta} \\ \\ \sf\pink \implies cot \theta\end{gathered}\end{gathered}$

$⟹ </p><p>sinθ(1+cosθ)</p><p>1+cosθ−sin 2 θ</p><p> </p><p> </p><p>sin 2θ=1−cos 2θ:$

$⟹ </p><p>sinθ(1+cosθ)</p><p>1+cosθ−(1−cos2θ)</p><p> </p><p> </p><p>(1−cos 2θ)=(1 2−cos 2 θ)=(1−cosθ)(1+cosθ):$

$⟹ </p><p>sinθ(1+cosθ)</p><p>1+cosθ−(1−cosθ)(1+cosθ)$

$⟹ </p><p>sinθ( 1+cos0)1+cosθ (1−(1−cosθ))$

$⟹ </p><p>sinθ</p><p>1−(1−cosθ)$

$⟹ </p><p>sinθ</p><p>1−</p><p>1+cosθ)$

$⟹ </p><p>sinθ</p><p>cosθ$

$⟹cot</p><p>θ$

RHS:

$\sf\green\implies cot \theta⟹cotθ$

$\therefore∴$

$\bold\red{LHS = RHS}$

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