Question

If cosec 0 - sin 0 = l and sec 0 - cos 0 = m, prove that + m² (l+ m² +3)=1 .
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Answer 1

HEY MATE YOUR ANSWER IS HERE.....

★ as the question says ★

( let theta be x )

$\cosec x \: - \sin \: x = l \\ and \: \\ \sec \: x \: - \cos \: x = m$

NOW ...

$cosec \: x \: - \: sin \: x \: = 1$

$\frac{1}{sin \: x} - sin \: x \: = l$

$\frac{1 - {sin}^{2}x }{sin \: x} = l$

$\frac{ {cos}^{2}x }{sin \: x} = l$

NOW ..

$sec \: x \: - \: \: cos \: x = m$

$\frac{1}{cos \: x} - \: cos \: x = m$

$\frac{1 - {cos}^{2}x }{cos \: x} = \: m$

$\frac{ {sin}^{2} x}{cos \: x} = \: m$

★ NOW TAKING L.H.S...★

l²m²( l²+m²+3)

$( \frac{ {cos}^{4}x }{ {sin}^{2} \: x})( \frac{ {sin}^{4} \: x }{ {cos}^{2} x}) (\frac{ {cos}^{4}x }{ {sin}^{2} \: x} + \frac{ {sin}^{4} \: x }{ {cos}^{2} x} \: + 3)$

$( {cos}^{2} x \: {sin}^{2} x)( \frac{ {cos}^{6} + {sin}^{6} \: + 3 {sin}^{2}x \: {cos}^{2}x }{ {sin}^{2} {cos}^{2} } )$

CUT SIN ²X COS²X .......

NOW ...

${cos}^{6} + {sin}^{6} \: + 3 {sin}^{2}x \: {cos}^{2}x$

can be written as....

$(( { {cos}^{2}x })^{3} + ({ {sin}^{2}x })^{3} )+ 3 {sin}^{2} x \: {cos}^{2}x$

USING IDENTITY....

a³+b³ = (a + b ) ( a²+ b² + ab )

$(( {cos}^{2} x \: + {sin}^{2} x)( {sin}^{4} x \: + {cos }^{4} x - {sin}^{2} x \: {cos}^{2} x)) + 3 {sin}^{2} x \: {cos}^{2} x$

NOW

sin²x + cos ² x = 1

THEN

$( { {sin}^{2}x) }^{2} + ( { {cos}^{2} }x)^{2} - {sin}^{2} x \: {cos}^{2} x) + 3 {sin}^{2} x \: {cos}^{2} x$

then

a²+b²= (a+b)²- 2ab

$( ({ {sin}^{2}x }+ { {cos}^{2} }x)^{2} - 2 {sin}^{2}x \: {cos}^{2} x) - {sin}^{2} x \: {cos}^{2} x) + 3 {sin}^{2} x \: {cos}^{2} x$

AGAIN SIN²+ COS²= 1

THEN ( SIN ²+COS²)² = 1² = 1

$( 1- 2 {sin}^{2}x \: {cos}^{2} x - {sin}^{2} x \: {cos}^{2} x) + 3 {sin}^{2} x \: {cos}^{2} x \:$

$1 - 3 {sin}^{2} x \: {cos}^{2} x + 3 {sin}^{2} x \: {cos}^{2} x \:$

= 1

L.H.S = R.H.S

THANKS FOR THE QUESTION...

HOPE IT HELPS

KEEP SMILING ☺️☺️☺️

Answer 2

Hope this helps you.

Thank you