Question

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

Answer 1

√1/sin-1 /1/sin+1+√1sin+1/1/sin-1
√1-sin/1+sin+√1+sin/1-sin
√1Squar-sin square /1²-sin²
√cos²/cos²
then square cancel root
cos/cos
1/sec/1/sec
sec+sec
2sec proved

Answer 2

Hey there !!

Prove that :- 

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

Solution :-

Solving LHS :-

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



•°• LHS = RHS .


✔✔ Hence, it is proved ✅✅.



THANKS



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