Question

Prove that (sin theta)/(1+cos theta) + (1+cos theta)/(sin theta)= 2cosec theta

​

Answer 1

To Prove :

$\\ \bf \: \frac{sin\theta}{1 + cos \theta} + \frac{1 + cos \theta}{sin \theta} = 2cosec \theta$

Solution :

In the given equation ,

• $\bf{\dfrac{sin\theta}{1+cos\theta}+\dfrac{1+cos\theta}{sin\theta}=LHS}$

• $\bf{2cosec\theta=RHS}$

First let us solve the LHS part ,

By taking LCM ,

$\\ \longrightarrow \bf \: \frac{(sin \theta)(sin \theta) + (1 + cos \theta)(1 + cos \theta)}{(1 + cos \theta)(sin \theta)}$

$\\ \longrightarrow \bf \: \frac{ {sin}^{2} \theta + {(1 + cos \theta)}^{2} }{sin \theta(1 + cos \theta)}$

using the identity ;

• (a+b)² = a² + b² + 2ab

$\\ \longrightarrow \bf \frac{ {sin }^{2} \theta + 1 + {cos}^{2} \theta + 2(1)(cos \theta ) }{(1 + cos \theta)sin \theta}$

$\\ \longrightarrow \bf \: \frac{ {sin}^{2} \theta + 1 + {cos}^{2} \theta + 2cos \theta }{(1 + cos \theta)sin \theta}$

We know that sin²θ + cos²θ = 1 ;

$\\ \longrightarrow \bf \: \frac{1 + 1 + 2cos \theta}{(1 + sin \theta)cos \theta}$

$\\ \longrightarrow \bf \: \frac{2 + 2cos \theta}{(1 + cos \theta)sin \theta}$

Taking 2 common in the numerator ;

$\\ \longrightarrow \bf \: \frac{2(1 + cos \theta)}{(1 + cos \theta)sin \theta}$

Cancelling (1 + cosθ) ;

$\\ \longrightarrow \bf \: \frac{2}{sin \theta}$

$\\ \longrightarrow \bf \: 2 \times \frac{1}{sin \theta}$

We know that ;

• $\\ \bf{\dfrac{1}{sin\theta}=cosec\theta}$

$\\ \longrightarrow \bf \: 2cosec \theta$

$\\ \longrightarrow \bf RHS$

Answer 2

Answer:

To Prove :

\begin{gathered} \\ \bf \: \frac{sin\theta}{1 + cos \theta} + \frac{1 + cos \theta}{sin \theta} = 2cosec \theta \\ \\ \end{gathered}

1+cosθ

sinθ

+

sinθ

1+cosθ

=2cosecθ

Solution :

In the given equation ,

\bf{\dfrac{sin\theta}{1+cos\theta}+\dfrac{1+cos\theta}{sin\theta}=LHS}

1+cosθ

sinθ

+

sinθ

1+cosθ

=LHS

\bf{2cosec\theta=RHS}2cosecθ=RHS

First let us solve the LHS part ,

By taking LCM ,

\begin{gathered} \\ \longrightarrow \bf \: \frac{(sin \theta)(sin \theta) + (1 + cos \theta)(1 + cos \theta)}{(1 + cos \theta)(sin \theta)} \\ \\ \end{gathered}

⟶

(1+cosθ)(sinθ)

(sinθ)(sinθ)+(1+cosθ)(1+cosθ)

\begin{gathered} \\ \longrightarrow \bf \: \frac{ {sin}^{2} \theta + {(1 + cos \theta)}^{2} }{sin \theta(1 + cos \theta)} \\ \\ \end{gathered}

⟶

sinθ(1+cosθ)

sin

2

θ+(1+cosθ)

2

using the identity ;

(a+b)² = a² + b² + 2ab

\begin{gathered} \\ \longrightarrow \bf \frac{ {sin }^{2} \theta + 1 + {cos}^{2} \theta + 2(1)(cos \theta ) }{(1 + cos \theta)sin \theta} \\ \\ \end{gathered}

⟶

(1+cosθ)sinθ

sin

2

θ+1+cos

2

θ+2(1)(cosθ)

\begin{gathered} \\ \longrightarrow \bf \: \frac{ {sin}^{2} \theta + 1 + {cos}^{2} \theta + 2cos \theta }{(1 + cos \theta)sin \theta} \\ \\ \end{gathered}

⟶

(1+cosθ)sinθ

sin

2

θ+1+cos

2

θ+2cosθ

We know that sin²θ + cos²θ = 1 ;

\begin{gathered} \\ \longrightarrow \bf \: \frac{1 + 1 + 2cos \theta}{(1 + sin \theta)cos \theta} \\ \\ \end{gathered}

⟶

(1+sinθ)cosθ

1+1+2cosθ

\begin{gathered} \\ \longrightarrow \bf \: \frac{2 + 2cos \theta}{(1 + cos \theta)sin \theta} \\ \\ \end{gathered}

⟶

(1+cosθ)sinθ

2+2cosθ

Taking 2 common in the numerator ;

\begin{gathered} \\ \longrightarrow \bf \: \frac{2(1 + cos \theta)}{(1 + cos \theta)sin \theta} \\ \\ \end{gathered}

⟶

(1+cosθ)sinθ

2(1+cosθ)

Cancelling (1 + cosθ) ;

\begin{gathered} \\ \longrightarrow \bf \: \frac{2}{sin \theta} \end{gathered}

⟶

sinθ

2

\begin{gathered} \\ \longrightarrow \bf \: 2 \times \frac{1}{sin \theta} \end{gathered}

⟶2×

sinθ

1

We know that ;

\begin{gathered}\\ \bf{\dfrac{1}{sin\theta}=cosec\theta}\end{gathered}

sinθ

1

=cosecθ

\begin{gathered} \\ \longrightarrow \bf \: 2cosec \theta \\ \\ \end{gathered}

⟶2cosecθ

\begin{gathered} \\ \longrightarrow \bf RHS \\ \\ \end{gathered}

⟶RHS