Question

kindly solve this...​

Answer 1

Answer

Given:

$\frac{ { \secØ }^{3} }{ { \sec }^{2}Ø - 1 } + \frac{ { \csc }^{3} Ø }{ { \csc }^{2}Ø - 1 } = \secØ \times \cscØ \times ( \secØ + \cscØ)$

$\colorbox{li} {we have, L.H.S.} = \frac{ { \sec }^{3}Ø }{ { \sec}^{2}Ø - 1 } + \frac{ { \csc }^{3} Ø}{ { \csc }^{2} - 1 }$

$⇒ \frac{ \frac{ { \sec }^{3}Ø }{ { \sec }^{2}Ø } }{ \frac{ { \sec }^{2} Ø - 1}{ { \sec}^{2} Ø} } + \frac{ \frac{ { \csc }^{3}Ø }{ { \csc }^{2} Ø} }{ \frac{ { \csc}^{2}Ø - 1 }{ { \csc}^{2} Ø} }$

$⇒ \secØ \: { \csc }^{2} + \cscØ \: { \sec }^{2} Ø$

$⇒ \secØ \: \cscØ( \secØ + \cscØ)$

$\colorbox{lighen} {R.H.S.}$

$\\ \\ \\ \\ \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙ⌨}$

Answer 2

Step-by-step explanation:

To prove,

$\boxed{\dfrac{sec^3\theta}{cosec^2\theta-1} + \dfrac{cosec^3\theta}{cosec^2\theta-1} = sec\theta cosec\theta (sec \theta + cosec \theta)}$

Solving LHS:

$\sf \bf :\longmapsto \dfrac{sec^3\theta}{sec^2\theta - 1} + \dfrac{cosec^3\theta}{cosec^2 \theta -1}$

$\sf \bf :\longmapsto \dfrac{sec^3 \theta}{tan^2\theta} + \dfrac{cosec^3 \theta}{cot^2 \theta}$

Converting into sin and cos,

$\sf \bf :\longmapsto \dfrac{1}{cos^3\theta} \times \dfrac{cos^2\theta}{sin^2 \theta} + \dfrac{1}{sin^3\theta} \times \dfrac{sin^2\theta}{cos^2\theta}$

$\sf \bf :\longmapsto \dfrac{1}{sin^2\theta cos\theta} + \dfrac{1}{sin\theta cos^2\theta}$

Converting back to sec and tan ,

$\sf \bf :\longmapsto cosec^2 \theta \ sec\theta + cosec \theta\ sec^2\theta$

$\sf \bf :\longmapsto sec\theta cosec\theta\ (sec\theta + cosec \theta)$

= RHS

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