Question

Give me answer. This question belong to trigonometry.​

Answer 1

After going through this you will get

sin2A + cos2A - 1 = 0

since sin2A + cos2A = 1

=> 1 - 1 = 0

Hence, proved.

Answer 2

GIVEN:

$x = \csc( \alpha ) + \cos( \alpha )$

$y = \csc( \alpha ) - \cos( \alpha )$

I LET A AS ALPHA.

TO PROVE :

$( { \frac{2}{x + y} })^{2} + ( { \frac{x - y}{2} })^{2}$

TRIGONOMETRIC IDENTITY USED :

${ \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1$

$\frac{1}{ \csc( \alpha ) } = \sin( \alpha )$

NOW WE HAVE TO FIND

$(x + y) \\ (x - y)$

$x + y = \csc( \alpha ) + \cos( \alpha ) + \csc( \alpha ) - \cos( \alpha ) \\ x + y = 2 \csc( \alpha )$

$x - y = \csc( \alpha ) + \cos( \alpha ) - \csc( \alpha ) + \cos( \alpha ) \\ ( sign \: will \: change \: here) \\ x - y = 2 \cos( \alpha )$

TAKING LHS :

$( { \frac{2}{ \csc( \alpha ) } })^{2} + ( { \frac{2 \cos( \alpha ) }{2} })^{2} - 1$

2 will cancelled out.

$( { \frac{1}{ \csc( \alpha ) } }^{2} + ( { \frac{2 \cos( \alpha ) }{1} }^{2} - 1 \\ ( \frac{1}{ \csc( \alpha ) } = \sin( \alpha )$

${ \sin( \alpha ) }^{2} + \cos( \alpha )^2 - 1$

$1 - 1$

$0$

LHS =RHS

HENCE PROVED