Question

Q12. If cosA + sinA = 1/2(cosA - sinA), then find sin²A.​

Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{0 \degree < A < 90\degree} \\ &\sf{cosA + sinA = \dfrac{1}{2(cosA - sinA)} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{ {sin}^{2}A}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \red{ \sf \: {cos}^{2}A = 1 - {sin}^{2}A}}$

$\boxed{ \red{ \sf \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:cosA + sinA = \dfrac{1}{2(cosA - sinA)}$

$\rm :\longmapsto\:2(cosA + sinA)(cosA - sinA) = 1$

$\rm :\longmapsto\:2( {cos}^{2}A - {sin}^{2}A) = 1$

$\rm :\longmapsto\:2(1 - {sin}^{2}A - {sin}^{2}A) = 1$

$\rm :\longmapsto\:2(1 - 2{sin}^{2}A) = 1$

$\rm :\longmapsto\:2 - 4{sin}^{2}A = 1$

$\rm :\longmapsto\:2 -1 = 4{sin}^{2}A$

$\rm :\longmapsto\:1 = 4{sin}^{2}A$

$\bf\implies \: {sin}^{2}A = \dfrac{1}{4}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1