Question

plz plz plz solve this sum ........​

Answer 1

To Prove:

$\sqrt{ \dfrac{1 + \cos A }{1 - \cos A} } + \sqrt{ \dfrac{ 1 - \cos A }{1 + \cos A } } = 2 \cosec A$

Method of Solution:

1. Don't Rationalize take L.C.M

2. Use Trigonometry Rules

3. Simplify and get Answer

Simple ?

Solution:

L.H.S

$\sqrt{ \dfrac{1 + \cos A }{1 - \cos A} } + \sqrt{ \dfrac{ 1 - \cos A}{1 + \cos A} } \\ \\ \frac{ \sqrt{(1 + \cos A)(1 + \cos A)} + \sqrt{(1 - \cos A)(1 - \cos A)}}{ \sqrt{(1 + \cos A)(1 - \cos A} )} \\ \\ = \frac{1 + \cos A+ 1 - \cos A}{ \sqrt{1 - { \cos}^{2} A } } \\ \\ = \frac{1 + 1}{ \sqrt{ \sin {}^{2}}} \\ \\ = \frac{2}{ \sin A} \\ \\ = 2.\cosec A \\ \\ \therefore \text{L.H.S = R.H.S} \: \:[\textsf{Proved}]$

Some important trigonometry Rules:

sinø . cosecø = 1

cosø . secø = 1

tanø . cotø = 1

sin²ø + cos²ø = 1

cosec²ø - cot² = 1

sec²ø - tan²ø = 1