Question

Guys! As we all know that . . . Sin^2 A + cos^2 A = 1 . . So , Sin^2 A = 1- cos^2 A Is it possible that Sin A = 1 - cos A if yes then prove it

Answer 1

Answer:

No

Step-by-step explanation:

No, It's impossible. The math masters can't prove this.

You can see

sina = 1-cosa a = 30*

sin30 = 1- cos 30

1/2 = 1-√3/2

it's not possible bro!!!!!!

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

Whether it is possible to Find A such that

$sinA = 1 - cosA$

EVALUATION

Yes it is possible to Find A such that

$sinA = 1 - cosA$

EXPLANATION

$sinA = 1 - cosA$

$\implies \: sinA + cosA = 1$

$\implies \: \displaystyle \: \dfrac{1}{ \sqrt{2} } sinA + \frac{1}{ \sqrt{2} } cosA = \frac{1}{ \sqrt{2} } \: \: .....(1)$

$\implies \: \displaystyle \: sinA \: cos \: {45}^{ \circ} + cosA \: cos \: {45}^{ \circ}: = sin \: {45}^{ \circ}$

$\implies \: \displaystyle \: sin(A + {45}^{ \circ} ) = sin {45}^{ \circ}$

$\implies \: \displaystyle \: (A + {45}^{ \circ} ) = {45}^{ \circ}$

$\implies \: A = {0}^{ \circ}$

Again from (1)

$\implies \: \displaystyle \: cosA \: cos \: {45}^{ \circ} + sinA \: cos \: {45}^{ \circ}: = cos \: {45}^{ \circ}$

$\implies \: \displaystyle \: cos(A \: - {45}^{ \circ} ) = cos \: {45}^{ \circ}$

$\implies \: \displaystyle \: (A \: - {45}^{ \circ} ) = \: {45}^{ \circ}$

$\implies \: \displaystyle \: A \ = cos \: {90}^{ \circ}$

RESULT

$\red {\fbox{THE REQUIRED VALUE OF A IS 0° , 90°}}$