Question

(1-sinA+cosA)(1-sinA+cosA)=2(1+cosA)(1-sinA)​

Answer 1

$\boxed{\begin{minipage}{7 cm}Fundamental Trigonometric Identities \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\1+\tan^2\theta = \sec^2\theta \\ \\1+\cot^2\theta = \text{cosec}^2 \, \theta \end{minipage}}$

$\mathfrak{Question:-}$

(1-sinA+cosA)(1-sinA+cosA)=2(1+cosA)(1-sinA)​

$\mathfrak{Solution:-}$

We take LHS part,

$\bold{=(1-sinA+cosA)(1-sinA+cosA)}$

$\bold{=(1 - sinA+cosA)^{2}}$

$\bold{=[(1-sinA)+(cosA)]^{2}}$

$\bold{= (1-sinA)^{2} + Cos^{2}A + 2(1-sinA)(cosA)}$

$\bold{=1-sin^{2}A-2sinA+1-sin^{2}A+2cosA-2sinAcosA}$

$\bold{=2-2sinA+2cosA-2sinAcosA}$

$\bold{=2(1-sinA)+2cosA(1-sinA)}$

$\bold{=(2+2cosA)(1-sinA)}$

$\bold{=2(1+cosA)(1-sinA)}$

$\bold{Henced\; Proved}$

Answer 2

$\huge\bf\pink{\mid{\overline{\underline{Your\: Answer}}}\mid}$

step-by-step explanation:

Given,

L.H.S = (1-sinA+cosA)(1-sinA+cosA)

= ${(1-sinA+cosA)}^{2}$

after arrangement of terms,

we get,

= ${[(1-sinA)+(cosA)]}^{2}$

now,

we know that,

${(a+b)}^{2}$ = ${a}^{2}$+ 2ab + ${b}^{2}$

so,

here,

a = (1-sinA)

b = cosA

putting these values in the identity,

we get,

= ${(1-sinA)}^{2}$+ 2(1-sinA)(cosA) + ${(cosA)}^{2}$

= { ${1}^{2}$ - 2(1)(sinA) +${(sinA)}^{2}$ } + 2(cosA)(1- sinA) + ${(cosA)}^{2}$

= 1 - 2sinA + ${sin}^{2}$A + 2(1-sinA)(cosA) + ${cos}^{2}$A

But,

we know that,

${sin}^{2}$A + ${cos}^{2}$A = 1

so putting this value,

we get,

= 1+ 1 + 2(1- sinA)(cosA) - 2 sinA

= 2+ 2(1- sinA) (cos A) - 2sinA

taking 2 as common from all terms,

we get,

= 2[ 1+ (1-sinA)(cosA) - sinA]

= 2[ ( 1-sinA) + (1-sinA)(cosA) ]

Now,

taking (1-sinA) as common term,

we get,

= 2(1-sinA)(1+cosA)

= 2( 1+cosA)(1- sinA)

= R.H.S

since,

L.H.S = R.H.S

hence,

proved..