Question

3. If cosec A - sin A =p and sec A-cos A = then prove that
(p²q)⅔ + (pq²)⅔ = 1
​

Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{cosecA - sinA = p} \\ &\sf{secA - cosA = q} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: prove - \begin{cases} &\sf{ {\bigg( { {p}^{2} q}\bigg) }^{\dfrac{2}{3} } +{\bigg( { {pq} }^{2} \bigg) }^{\dfrac{2}{3}}= 1}\end{cases}\end{gathered}\end{gathered}$

Identities Used :-

$\boxed{ \red{ \bf \: cosecx = \dfrac{1}{sinx}}}$

$\boxed{ \red{ \bf \: secx = \dfrac{1}{cosx}}}$

$\boxed{ \red{ \bf \: {sin}^{2}x + {cos}^{2}x = 1}}$

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

Given that

$\rm :\longmapsto\:cosecA - sinA = p$

$\rm :\longmapsto\:\dfrac{1}{sinA} - sinA = p$

$\rm :\longmapsto\:\dfrac{1 - {sin}^{2}A }{sinA} = p$

$\bf\implies \boxed{ \bf \:p = \dfrac{ {cos}^{2} A}{sinA}} - - - (1)$

Again,

$\rm :\longmapsto\:secA - cosA = q$

$\rm :\longmapsto\:\dfrac{1}{cosA} - cosA = q$

$\rm :\longmapsto\:\dfrac{1 - {cos}^{2}A }{cosA} = q$

$\bf\implies \boxed{ \bf \:q = \dfrac{ {sin}^{2} A}{cosA}} - - - (2)$

Now,

Consider LHS,

$\rm :\longmapsto\:{\bigg( {p}^{2}q \bigg) }^{\dfrac{2}{3} } + {\bigg( {q}^{2}p \bigg) }^{\dfrac{2}{3} }$

On substituting the values of p and q from equation (1) and equation (2), we get

$\rm :\longmapsto\: = {\bigg(\dfrac{ {cos}^{4}A }{ {sin}^{2} A} \times \dfrac{ {sin}^{2}A }{cosA} \bigg) }^{\dfrac{2}{3} } + {\bigg(\dfrac{ {sin}^{4}A }{ {cos}^{2} A} \times \dfrac{ {cos}^{2}A }{sinA} \bigg) }^{\dfrac{2}{3} }$

$\rm :\longmapsto\: \: = \: \:{\bigg( {cos}^{3} A \bigg) }^{\dfrac{2}{3}} + {\bigg( {sin}^{3}A\bigg) }^{\dfrac{2}{3} }$

$\rm :\longmapsto\: \: = \: \: {cos}^{2} A + {sin}^{2} A$

$\rm :\longmapsto\: \: = \: \:1$

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

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