Question

The value of (sinθ+cosθ)²- 2sin(90°-θ) cos(90°-θ) is equal to

Answer 1

here...is the answer....

i hope it helps....you....

Answer 2

QUESTION:

I LET THETHA AS ALPHA.

THE VALUE OF :

${( \sin( \alpha ) + \cos( \alpha ) )}^{2} - 2 \sin(90 - \alpha ) \times \cos(90 - \alpha )$

WE USE HERE TRIGONOMETRIC IDENTITY :

1.

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

2.

$\sin(90 - \alpha ) = \cos( \alpha )$

3.

$\cos(90 - \alpha ) = \sin( \alpha )$

OTHER identity we used is :

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

NOW, ANSWER :

${( \sin( \alpha ) + \cos( \alpha ) }^{2} - 2 \sin(90 - \alpha ) \cos(90 - \alpha )$

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

$1 + 2 \sin( \alpha ) \times \cos( \alpha ) - 2 \sin( \alpha ) \times \cos( \alpha )$

$</u></em></strong><strong><em><u>\</u></em></strong><strong><em><u>red</u></em></strong><strong><em><u>{</u></em></strong><strong><em><u>2 \sin( \alpha ) \times \cos( \alpha ) \: will \: be \: cancelled \: out</u></em></strong><strong><em><u>}</u></em></strong><strong><em><u>$

$= 1$

SO FINAL ANSWER IS 1.