Question

solve this problem fast

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Answer 1

$\huge \red{\mathfrak{solution}}$

Given:

$\star \: \cosec \theta = 2$

To show :

$\red\star \: \cot \theta + \frac{ \sin \theta}{1 + \cos \theta} = 2$

LHS

$\rightarrow \: \frac{ \cos \theta}{ \sin \theta} + \frac{sin \theta}{1 + cos \theta} \\ \\ \rightarrow \: \frac{ \cos \theta + { \cos}^{2} \theta + { \sin}^{2} \theta}{ \sin \theta(1 + \cos \theta)} \\ \\ \rightarrow \: \frac{ \cancel{1 + \cos \theta}}{ \sin \theta \cancel{(1 + \cos \theta)} } \\ \\ \rightarrow \: \frac{1}{ \sin \theta} \\ \\ \rightarrow \: \cosec \theta \\ \\ \rightarrow \: 2$

LHS = RHS

Formula :

• cot@= cos@/sin@

•sin^2@+cos^2@= 1

•sin@=1/cosec@

$\huge \boxed{ \mathfrak{ \green{proved}}}$

Answer 2

Step-by-step explanation:

Given:

$\star \: \cosec \theta = 2⋆cosecθ=2$

To show :

$\begin{lgathered}\red\star \: \cot \theta + \frac{ \sin \theta}{1 + \cos \theta} = 2 \\\end{lgathered}⋆cotθ+1+cosθsinθ=2$

LHS

$\begin{lgathered}\rightarrow \: \frac{ \cos \theta}{ \sin \theta} + \frac{sin \theta}{1 + cos \theta} \\ \\ \rightarrow \: \frac{ \cos \theta + { \cos}^{2} \theta + { \sin}^{2} \theta}{ \sin \theta(1 + \cos \theta)} \\ \\ \rightarrow \: \frac{ \cancel{1 + \cos \theta}}{ \sin \theta \cancel{(1 + \cos \theta)} } \\ \\ \rightarrow \: \frac{1}{ \sin \theta} \\ \\ \rightarrow \: \cosec \theta \\ \\ \rightarrow \: 2\end{lgathered}→sinθcosθ+1+cosθsinθ→sinθ(1+cosθ)cosθ+cos2θ+sin2θ→sinθ(1+cosθ)1+cosθ→sinθ1→cosecθ→2$

LHS = RHS