Question

heyaa!! all maths experts this side...

Answer 1

Easy.
wait solving. Hi I am Shinde , Om Shinde.

Answer 2

SOLUTION :-

${a}^{2} { \sec}^{2} \alpha \: - {b}^{2} { \tan }^{2} \alpha \: = { \: c}^{2}$
we know that :-

$\sec( \alpha ) = \frac{1}{ \cos( \alpha ) } \\ \\ and \\ \\ \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

so,


${a}^{2} \frac{1}{ { \cos }^{2} \alpha } - {b}^{2} \frac{ \ { \sin}^{2} \alpha }{ { \cos }^{2} \alpha } = {c}^{2}$

$\frac{ {a}^{2} - {b}^{2} \ { \sin }^{2} \alpha }{ { \cos}^{2} \alpha } = {c}^{2}$

Now , we know that :-

☞ cos^2 alpha = 1 - sin^2 alpha.....


$\frac{ {a}^{2} - {b}^{2} { \sin}^{2} \alpha }{1 - { \sin }^{2} \alpha } = {c}^{2}$


Now,

${a}^{2} - {b}^{2} { \sin }^{2} \alpha = {c}^{2} (1 - { \sin}^{2} \alpha )$


${a}^{2} - {b}^{2} { \sin}^{2} \alpha = {c}^{2} - {c}^{2} { \sin}^{2} \alpha \: \:$


${c}^{2} { \sin }^{2} \alpha \: - \: {b}^{2} { \sin}^{2} \alpha = {c}^{2} - {a}^{2}$


${ \sin}^{2} \alpha ( {c}^{2} - {b}^{2} ) = {c}^{2} - {a}^{2}$


${ \sin}^{2} \alpha = \frac{ {c }^{2} - {a}^{2} }{ {c}^{2} - {b}^{2} }$


$\sin \alpha = + \sqrt{ \frac{ {c}^{2} - {a}^{2} }{ {c }^{2} - {b}^{2} } } \: \: \: or \: \: \: - \sqrt{ \frac{ {c}^{2} - {a}^{2} }{ {c}^{2} - {b}^{2} } }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{hence.........proved}{hope \: this \: will \: help \: you \: } \: \: \: \: \: \: \: \: \: \: \:$





☛ ⛧⛧ By, Ⓜr. Thakur ⛧⛧