Question

please solve it friends..... ​

Answer 1

Step-by-step explanation:

Solution:

We have to prove:-

\frac{\cot \theta+\csc \theta-1}{\cot \theta-\csc \theta+1}=\frac{1+\cos \theta}{\sin \theta}

cotθ−cscθ+1

cotθ+cscθ−1

=

sinθ

1+cosθ

Using trigonometric identity:

\csc ^{2} \theta-\cot ^{2} \theta=1csc

2

θ−cot

2

θ=1

Applying this in L.H.S of equation we get,

\frac{\cot \theta+\csc \theta-(\csc ^{2} \theta-\cot ^{2} \theta)}{\cot \theta-\csc \theta+1}

cotθ−cscθ+1

cotθ+cscθ−(csc

2

θ−cot

2

θ)

\text { Using } a^{2}-b^{2}=(a+b)(a-b), \text { we get } Using a

2

−b

2

=(a+b)(a−b), we get

\frac{\cot \theta+\csc \theta-(\csc \theta+\cot \theta)(\csc \theta-\cot \theta)}{\cot \theta-\csc \theta+1}

cotθ−cscθ+1

cotθ+cscθ−(cscθ+cotθ)(cscθ−cotθ)

\frac{(\cot \theta+\csc \theta)\{1-(\csc \theta-\cot \theta)\}}{\cot \theta-\csc \theta+1}

cotθ−cscθ+1

(cotθ+cscθ){1−(cscθ−cotθ)}

\text {on taking } \cot \theta+\csc \theta \text { as common }on taking cotθ+cscθ as common

\frac{(\cot \theta+\csc \theta)\{1-\csc \theta+\cot \theta\}}{\cot \theta-\csc \theta+1}

cotθ−cscθ+1

(cotθ+cscθ){1−cscθ+cotθ}

Cancelling terms we get,

\frac{(\cot \theta+\csc \theta)\{1-\csc \theta+\cot \theta\}}{\cot \theta-\csc \theta+1}

cotθ−cscθ+1

(cotθ+cscθ){1−cscθ+cotθ}

=\cot \theta+\csc \theta=\frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta}=\frac{\cos \theta+1}{\sin \theta}=\mathrm{R.H.S}=cotθ+cscθ=

sinθ

cosθ

+

sinθ

1

=

sinθ

cosθ+1

=R.H.S

Hence proved

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