Question

Prove that:

$(1-\sin A+\cos A{)}^2=2(1+\cos A\left)\right(1-\sin A)$

Answer 1

To prove: $(1-\sin A+\cos A{)}^2=2(1+\cos A\left)\right(1-\sin A)$

Lets take LHS and then equate it to RHS.

LHS $=(1-\sin A+\cos A{)}^2$

$=\left(\right(1-\sin A)+\cos A{)}^2$

$=(1-\sin A{)}^2+{\cos }^{2}A+2(1-\sin A\left)\right(\cos A)$

[$\because (a+b{)}^2=a^2+b^2+2ab$]

$=1+{\sin }^{2}A-2\sin A+{\cos }^{2}A+2(1-\sin A)\left(\cos A\right)$

$=1+{\sin }^{2}A+{\cos }^{2}A-2\sin A+2(1-\sin A)\left(\cos A\right)$

$=1+1-2\sin A+2(1-\sin A)\left(\cos A\right)$ [$\because {\sin }^{2}A+{\cos }^{2}A=1$]

$=2-2\sin A+2(1-\sin A)\left(\cos A\right)$

$=2(1-\sin A)+2(1-\sin A)\left(\cos A\right)$

$=2(1-\sin A)(1+\cos A)$

$=$ RHS

Therefore, LHS $=$ RHS.

Hence, $(1-\sin A+\cos A{)}^2=2(1+\cos A\left)\right(1-\sin A)$ proved.