Question

If cosQ+sinQ=√2,then cosQ-sinQ is equal to???

Answer 1

$\cos( \beta ) + \sin( \beta ) = \sqrt{2} \\ \\ { \big( \cos( \beta ) + \sin( \beta ) \big)}^{2} = {( \sqrt{2}) }^{2} \\ \\ { \sin( \beta ) }^{2} + { \cos( \beta ) }^{2} + 2 \sin( \beta ) \cos( \beta ) = 2 \\ \\ 1 + 2 \sin( \beta ) \cos( \beta ) = 2 \\ \\ 2 \sin( \beta ) \cos( \beta ) = 2 - 1 = 1$

__________________

$\cos( \beta ) - \sin( \beta ) \\ \\ = \sqrt{ \big( { \cos( \beta ) - \sin( \beta ) \big )}^{2} } \\ \\ = \sqrt{ { \cos( \beta ) }^{2} + { \sin( \beta ) }^{2} - 2 \sin( \beta ) \cos( \beta ) } \\ \\ = \sqrt{1 - 2 \sin( \beta ) \cos( \beta ) } \\ \\ = \sqrt{1 - 1} \\ \\ = 0$

Answer : cosQ-sinQ = 0

Note: I have used beta in place of Q.

Answer 2

Answer:

→ cosQ-sinQ = 0

Step-by-step explanation:

Note: I am going to use beta in place of Q.

$\begin{lgathered}\cos( \beta ) + \sin( \beta ) = \sqrt{2} \\ \\ { \big( \cos( \beta ) + \sin( \beta ) \big)}^{2} = {( \sqrt{2}) }^{2} \\ \\ { \sin( \beta ) }^{2} + { \cos( \beta ) }^{2} + 2 \sin( \beta ) \cos( \beta ) = 2 \\ \\ 1 + 2 \sin( \beta ) \cos( \beta ) = 2 \\ \\ 2 \sin( \beta ) \cos( \beta ) = 2 - 1 = 1\end{lgathered}$

$\begin{lgathered}\cos( \beta ) - \sin( \beta ) \\ \\ = \sqrt{ \big( { \cos( \beta ) - \sin( \beta ) \big )}^{2} } \\ \\ = \sqrt{ { \cos( \beta ) }^{2} + { \sin( \beta ) }^{2} - 2 \sin( \beta ) \cos( \beta ) } \\ \\ = \sqrt{1 - 2 \sin( \beta ) \cos( \beta ) } \\ \\ = \sqrt{1 - 1} \\ \\ = 0\end{lgathered}$