Question

Prove that under root Coseca -1/CosecA +1+under root cosec A+1/cosec A-1 = 2 sec A

Answer 1

Hey there !!

Prove that :-

$\sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} } + \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1} } = 2 \sec \theta.$

Solution :-

Solving LHS :-

$= \sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} } + \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1} } \\ \\ = \sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} \times \frac{ \cosec \theta - 1}{\cosec \theta - 1} } + \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1} \times \frac{\cosec \theta + 1}{\cosec \theta + 1} } . \\ \\ = \sqrt{ \frac{ {(\cosec \theta - 1)}^{2} }{ {\cosec}^{2} \theta - {1}^{2} } } + \sqrt{ \frac{ {(\cosec \theta + 1)}^{2} }{ {\cosec}^{2} \theta - {1}^{2} } } . \\ \\ = \sqrt{ \frac{ {(\cosec \theta - 1)}^{2} }{ { - \cot}^{2} \theta} } + \sqrt{ \frac{ {(\cosec \theta + 1)}^{2} }{ { - \cot}^{2} \theta} } . \\ \\ = \frac{ \cosec\theta - 1}{ \cot\theta} + \frac{ \cosec\theta + 1}{ \cot\theta} . \\ \\ = \frac{ \cosec\theta }{ \cot\theta} - \cancel\frac{1}{ \cot\theta} + \frac{ \cosec\theta}{ \cot\theta} + \cancel\frac{1}{ \cot\theta} . \\ \\ = \frac{ \frac{1}{ \cancel{\sin\theta} }}{ \frac{ \cos\theta}{ \cancel{\sin\theta} }} + \frac{ \frac{1}{ \cancel{\sin\theta} }}{ \frac{ \cos\theta}{ \cancel{\sin\theta} }} . \\ \\ = \frac{1}{ \cos\theta} + \frac{1}{ \cos \theta} . \\ \\ = \sec \theta + \sec \theta. \\ \\ = \boxed{ \boxed{ \pink{ = 2 \sec \theta.}}} \checkmark \checkmark.$



•°• LHS = RHS .


✔✔ Hence, it is proved ✅✅.



THANKS



#BeBrainly.

Answer 2

Answer:

To prove :$\sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} } +  \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1} }  = 2 \sec \theta.$

Solution :-

$=  \sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} } +  \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1} }$

$=  \sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} \times  \frac{ \cosec \theta  -  1}{\cosec \theta  -  1}  } +  \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1}  \times  \frac{\cosec \theta + 1}{\cosec \theta + 1} } .$

$=  \sqrt{ \frac{ {(\cosec \theta - 1)}^{2} }{ {\cosec}^{2}  \theta -  {1}^{2}  }  }  + \sqrt{ \frac{ {(\cosec \theta  + 1)}^{2} }{ {\cosec}^{2}  \theta -  {1}^{2}  }  }$

$=  \sqrt{ \frac{ {(\cosec \theta - 1)}^{2} }{ {  - \cot}^{2}  \theta} }  + \sqrt{ \frac{ {(\cosec \theta  +  1)}^{2} }{ {  - \cot}^{2}  \theta} } .$

$= \frac{ \cosec\theta - 1}{ \cot\theta}  +   \frac{ \cosec\theta  +  1}{ \cot\theta} .$

$=  \frac{ \cosec\theta }{ \cot\theta}  -   \cancel\frac{1}{ \cot\theta}  +  \frac{ \cosec\theta}{ \cot\theta}  +   \cancel\frac{1}{ \cot\theta} .$

$=  \frac{ \frac{1}{  \cancel{\sin\theta} }}{ \frac{ \cos\theta}{  \cancel{\sin\theta} }}  + \frac{ \frac{1}{  \cancel{\sin\theta} }}{ \frac{ \cos\theta}{  \cancel{\sin\theta} }}$

$=  \frac{1}{ \cos\theta}  +  \frac{1}{ \cos \theta}$

$=  \sec \theta  +  \sec \theta.$

$=   2 \sec \theta$

Hence proved