Question

PROOVE THIS TRIGONOMETRY!

Answer 1

hope it may help you.........

Answer 2

Question:

To Prove :

${\sf{ {\sqrt{ {\dfrac{1 - cos A}{1 + cos A}} }} = {\dfrac{sin A}{1 + cos A}} }}$

Step-by-step explanation:

L.H.S. = ${\sf{\ \ {\sqrt{ {\dfrac{1 - cos A}{1 + cos A}} }}}}$

_____________________________

Multiplying and dividing with (1 + cos A).

$\implies{\sf{ {\sqrt{ {\dfrac{1 - cos A}{1 + cos A}} \times {\dfrac{1 + cos A}{1 + cos A}} }} }}$

_____________________________

$\implies{\sf{ {\sqrt{ {\dfrac{(1 - cos A)(1 + cos A)}{(1 + cos A)^2}}}}}}$

_____________________________

${\boxed{\tt{Identity \ : \ (a - b)(a + b) = a^2 - b^2}}}$

${\tt{Here, \ a = 1, \ b = cos A}}$

_____________________________

$\implies{\sf{ {\sqrt{ {\dfrac{(1)^2 - (cos A)^2}{(1 + cos A)^2}} }}}}$

_____________________________

$\implies{\sf{ {\sqrt{ {\dfrac{1 - cos^2 A}{(1 + cos A)^2}}}}}}$

_____________________________

${\boxed{\tt{Identity \ : \ sin^2 \theta + cos^2 \theta = 1}}}$

${\tt{From \ this, \ we \ get \ [ 1 - cos^2 \theta = sin^2 \theta ] }}$

_____________________________

$\implies{\sf{ {\sqrt{ {\dfrac{sin^2 A}{(1 + cos A)^2}}}}}}$

_____________________________

We can write this as :

$\implies{\sf{ {\sqrt{ {\dfrac{(sin A)^2}{(1 + cos A)^2}}}}}}$

_____________________________

$\implies{\sf{ {\sqrt{ \left( {\dfrac{sin A}{1 + cos A}} \right) ^2 }}}}$

_____________________________

$\implies{\sf{ {\dfrac{sin A}{1 + cos A}}}}$

_____________________________

= R.H.S.

Hence, proved !!