Question

prove that

1+ sin A
cosec A - cot A
1-sin A
cosec A+cot A
= 2(1 + cot A)​

Answer 1

Answer:

your answer attached in the photo

Answer 2

Solution →

$\frac{1 + sin \: a}{cosec \: a \: - cot \: a} - \frac{1 - sin \: a}{cosec \: a \: + \: cot \: a} = 2(1 + cot \: a)$

Left hand side →

$= \frac{1 + sin \: a}{cosec \: a \: - cot \: a} - \frac{1 - sin \: a}{cosec \: a \: + \: cot \: a}$

We know that →

• $cosec \: a \: = \frac{1}{sin \: a}$
• $cot \: a \: = \frac{cos \: a}{sin \: a}$

Put the values of (cosec a) and (cot a) →

$= \frac{1 + sin \: a}{ \frac{1}{sin \: a} - \frac{cos \: a}{sina} } - \frac{1 - sin \: a}{ \frac{1}{sin \: a } + \frac{cos \: a}{sin \: a} }$

$= \frac{1 + sin \: a}{ \frac{1 - cos \: a}{sin \: a} } - \frac{1 - sin \: a}{ \frac{1 + cos \: a}{sin \: a} }$

$= \frac{sin \: a(1 + sin \: a)}{1 - cos \: a} - \frac{sin \: a(1 - sin \: a)}{1 + cos \: a}$

Let's take common (sin a) →

$= sin \: a ( \frac{1 + sin \: a}{1 - cos \: a} - \frac{1 - sin \: a}{1 + cos \: a} )$

$= sin \: a( \frac{(1 + cos \: a)(1 + sin \: a) - (1 - cos \: a)(1 - sin \: a)}{(1 - cos \: a)(1 + cos \: a)</strong><strong>)</strong><strong>}$

We know that →

• (a-b)(a+b) = a²-b²

Like this →

• (1- cos a)(1+cos a) = 1²-cos ² a
• = 1 - cos²a

$= sin \: a( \frac{(1 + cos \: a)(1 + sin \: a) - (1 - cos \: a)(1 - sin \: a)}{1 - {cos}^{2}a } )$

Now →

• 1 - cos² a = Sin²a

$= sin \: a( \frac{(1 + cos \: a)(1 + sin \: a) - (1 - cos \: a)(1 - sin \: a)}{ {sin}^{2}a } )$

$= \frac{sin \: a}{ {sin}^{2}a } (1 + sin \: a + cos \: a + cos \: a \times sin \: a - 1 + sin \: a + cos \: a - cos \: a \times sin \: a)$

Here ,

• Sin a will be cancelled.
• 1 cancelled by 1.
• cos a × sin a will be cancelled by cos a × sin a.

$= \frac{1}{sin \: a} (2sin \: a \: + \: 2 \: cos \: a)$

$</strong><strong>=</strong><strong> \frac{2 \: sin \: a}{sin \: a} + \: \frac{2 \: cos \: a}{sin \: a}$

(Sin a) will be cancelled by (sin a) →

$= 2 + 2 \: cot \: a$

$\boxed{ = 2(1 + cot \: a)}$ RHS

Hence proved !!

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