Question

prove that this question mate ​

Answer 1

Answer:

answer for the given problem is given

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Given :

$tan \theta + sin \theta = m$

$tan \theta - sin \theta = n$

To prove :

${ {(m}^{2} - {n}^{2}) }^{2} = 16mn$

Taking L.H.S:-

Note :-[We take Alpha

instead of theta ]

${m}^{2} - {n}^{2} = (tan \alpha + sin \alpha + tan \alpha - sin \alpha )(tan \alpha + sin \alpha - tan \alpha + sin \alpha )$

$= > {m}^{2} - {n}^{2} = (2tan \alpha )(2sin \alpha )$

$= > {m}^{2} - {n}^{2} = 4tan \alpha sin \alpha$

$\bold{{ {(m}^{2} - {n}^{2} ) }^{2} = 16 {tan}^{2} \alpha {sin}^{2} \alpha ....L.H.S}$

R.H.S=16mn

$= > 16(tan \alpha + sin \alpha )(tan \alpha - sin \alpha )$

$= > 16( {tan}^{2} \alpha - {sin}^{2} \alpha )$

$= > 16[\frac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } - {sin}^{2} \alpha ]$

$= > 16[\frac{ {sin}^{2} \alpha - {sin}^{2} \alpha {cos}^{2} \alpha }{ {cos}^{2} \alpha } ]$

$= > 16[ \frac{ {sin}^{2} \alpha (1 - {cos }^{2} \alpha )}{ {cos}^{2} \alpha } ]$

$\bold{\orange{\boxed{ {sin}^{2} \alpha + {cos}^{2} \alpha = 1}}}$

$= > 16[ \frac{ {sin}^{2} \alpha \times {sin}^{2} \alpha }{ {cos}^{2} \alpha } ]$

$= > 16[ \frac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } \times {sin}^{2} \alpha ]$

$\bold{ = > 16 {tan}^{2} \alpha {sin}^{2} \alpha}$

∴L.H.S=R.H.S