Question

Prove the following identity :

(1) sin theta / 1 + cos theta + 1 + cos theta / sin theta = 2cosec theta​

Answer 1

`\large\bf{\underline{\underline{\mathfrak{Answer}:}}}`

The following identity can be proved.

`\large\bf{\underline{\underline{\mathfrak{Step\:by\:step\:explanation}:}}}`

To prove the identities we need to choose more complex side first either L.H.S. or R.H.S. of the equation this is because it makes it easier to make further simplifications to prove the identity, then try to simplify it as much as possible (with a logical approach).

If the simplification is logical and equal to the other side of the equation then it is proved otherwise take the other side of the equation and try to prove it from the same approach.

Here,

L.H.S -

`:{\Longrightarrow{\large{\rm{(Sin\:\theta)/(1\:+\:Cos\:\theta)+(1\:+\:Cos\:\theta)/(Sin\:\theta)}}}}`

`:{\Longrightarrow{\large{\rm{(Sin^2\:\theta\:+\:(1\:+\:Cos\:\theta)^2)/(Sin\:\theta(1\:+\:Cos\:\theta))}}}}`

We know that,

`:{\Longrightarrow{\boxed{\small{\rm{(a\:+\:b)^2=a^2\:+\:b^2\:+\:2ab}}}}}`

Here in this question,

`:{\Longrightarrow{\small{\rm{a=1}}}}`

`:{\Longrightarrow{\small{\rm{b=Cos\:\theta}}}}`

Now,

`:{\Longrightarrow{\large{\rm{(Sin^2\:\theta\:+\:(1\:+\:Cos^2\:\theta\:+\:2\:Cos\:\theta))/(Sin\:\theta(1\:+\:Cos\:\theta))}}}}`

Next,

`:{\Longrightarrow{\large{\rm{((Sin^2\:\theta\:+\:Cos^2\:\theta)\:+\:1\:+\:2\:Cos\:\theta)/(Sin\:\theta(1\:+\:Cos\:\theta))}}}}`

We also know that,

`:{\Longrightarrow{\boxed{\small{\rm{Sin^2\:\theta\:+\:Cos^2\:\theta=1}}}}}`

So,

`:{\Longrightarrow{\large{\rm{(1\:+\:1\:+\:2\:Cos\:\theta)/(Sin\:\theta(1\:+\:Cos\:\theta))}}}}`

`:{\Longrightarrow{\large{\rm{(2\:+\:2\:Cos\:\theta)/(Sin\:\theta(1\:+\:Cos\:\theta))}}}}`

`:{\Longrightarrow{\large{\rm{(2(1\:+\:Cos\:\theta))/(Sin\:\theta(1\:+\:Cos\:\theta))}}}}`

`:{\Longrightarrow{\large{\rm{\frac{2\cancel{(1\:+\:Cos\:\theta)}}{Sin\:\theta\cancel{(1\:+\:Cos\:\theta)}}}}}}`

`:{\Longrightarrow{\large{\rm{(2)/(Sin\:\theta)}}}}`

`:{\Longrightarrow{\small{\rm{2Cosec\:\theta}}}}`

`{\therefore{\small{\rm{L.H.S=R.H.S}}}}`

Answer 2

See below for proof.

Explanation:

Given trigonometric identity:

`(\sin \theta)/(1 + \cos \theta) + (1 + \cos \theta)/(\sin \theta) = 2\csc \theta`

To prove the given identity, manipulate the left side of the equation until it matches the right side.

Find a common denominator to add the two fractions:

`(\sin \theta)/(1 + \cos \theta) \cdot (\sin \theta)/(\sin \theta) + (1 + \cos \theta)/(\sin \theta) \cdot (1 + \cos \theta)/(1 + \cos \theta)= 2\csc \theta`

Simplify each fraction:

`(\sin^2 \theta)/(\sin \theta (1 + \cos \theta)) + ((1 + \cos \theta)^2)/(\sin \theta (1 + \cos \theta))= 2\csc \theta`

Combine the fractions:

`(\sin^2 \theta + (1 + \cos \theta)^2)/(\sin \theta (1 + \cos \theta))= 2\csc \theta`

Expand (1 + cos θ)² in the numerator:

`(\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta)/(\sin \theta (1 + \cos \theta))=2\csc\theta`

Use the Pythagorean trigonometric identity, sin²θ + cos²θ = 1:

`(1 + 1 + 2\cos \theta)/(\sin \theta (1 + \cos \theta))=2\csc\theta`

Simplify the numerator:

`(2 + 2\cos \theta)/(\sin \theta (1 + \cos \theta))=2\csc\theta`

Factor out 2 from the numerator:

`(2(1 + \cos \theta))/(\sin \theta (1 + \cos \theta))=2\csc\theta`

Cancel out the common factor of (1 + cos θ) in the numerator and denominator:

`(2)/(\sin \theta)=2\csc\theta`

Use the cosecant reciprocal identity, 1 / sin θ = cosec θ:

`2\csc \theta=2\csc\theta`

Therefore, we have successfully proven the given identity:

`(\sin \theta)/(1 + \cos \theta) + (1 + \cos \theta)/(\sin \theta) = 2\csc \theta`