Question

(cos theta-2cos^3 theta)/2 (sin^3 theta- sin theta)= cot theta
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Answer 1

Correct Statement is

$\bf \: Prove \: that \sf \: \dfrac{cos \theta \: - 2 {cos}^{3}\theta \:}{2 {sin}^{3}\theta \: - sin\theta \: } = cot\theta \:$

Answer

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

$1. \: \: \: \boxed{ \bf \: {sin}^{2} \theta \: + {cos}^{2} \theta \: = 1}$

$2. \: \: \: \boxed{ \bf \dfrac{cos\theta \:}{sin\theta \:} = cot\theta \:}$

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

Consider,

$\rm :\longmapsto\:\dfrac{cos \theta \: - 2 {cos}^{3}\theta \:}{2 {sin}^{3}\theta \: - sin\theta \: }$

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

$\rm :\longmapsto\: = \: cot\theta \: \times \dfrac{ {sin}^{2}\theta \: + {cos}^{2}\theta \: - {2cos}^{2}\theta \:}{ {2sin}^{2}\theta \: - {sin}^{2}\theta \: - {cos}^{2}\theta \:}$

$\rm :\longmapsto\: = \: cot\theta \: \times \dfrac{ \cancel{ {sin}^{2}\theta \: - {cos}^{2}\theta} \:}{ \cancel{{sin}^{2}\theta \: - {cos}^{2}\theta} \:}$

$\rm :\longmapsto\: = \: cot\theta \:$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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