Question

If (a/cosec2)+(b/sec2)=c, prove that, cot2=(c-a)/(b-c)

Answer 1

Answer refer to the attachment

Answer 2

Given :-

$\tt \:  \dfrac{a}{ {cosec }^{2} A} + \dfrac{b}{ {sec }^{2} A} = c$

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To prove :-

$\tt \:  {cot}^{2} A = \dfrac{c - a}{b - c}$

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

$1. \boxed{ \blue{ \bf \:cosA = \dfrac{1}{secA} }}$

$2. \boxed{ \blue{ \bf \:sinA \: = \dfrac{1}{cosec A} }}$

$3. \boxed{ \blue{ \bf \: {cosec }^{2} = 1 + {cot}^{2} A}}$

$4. \boxed{ \blue{ \bf \:\dfrac{cosA}{sinA} = cotA}}$

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$\large\underline\purple{\bold{Solution :- }}$

Consider,

$\tt \:  \dfrac{a}{ {cosec }^{2} A} + \dfrac{b}{ {sec }^{2} A} = c$

$\tt \:  :\implies a {sin}^{2} A + b {cos}^{2} A = c$

$\red{ \bf \: Divide \: both \: sides \: by \: {sin}^{2} A \: we \: get}$

$\tt \: :\implies a \: + \: b \: {cot}^{2} A = c \: {cosec }^{2} A$

$\tt \: :\implies a \: + \: b \: {cot}^{2} A = c \: (1 + {cot }^{2} A)$

$\tt \: :\implies a \: + \: b \: {cot}^{2} A = c \: + \: c \: {cot }^{2} A$

$\tt \::\implies b \: {cot}^{2} A - c \: {cot}^{2} A = c \: - \: a$

$\tt \::\implies (b \: - \: c) {cot}^{2} A = c \: - \: a$

$\bf\implies \: {cot}^{2} A = \dfrac{c - a}{b - c}$

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

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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