Question

PLZ PROVE THIS QUESTION

I WILL DEFINITELY MARK AS BRAINLIEST....PLZ ......


DON'T GIVE NON-USEFUL ANS......PLZ​

Answer 1

QUESTION:

Prove that

$(1 + \frac{1}{ { \tan}^{2} \theta })(1 + \frac{1}{ { \cot}^{2} \theta } ) = \frac{ 1 }{ { \sin}^{2} \theta - {\sin}^{4} \theta }$

TO PROVE:

$(1 + \frac{1}{ { \tan}^{2} \theta })(1 + \frac{1}{ { \cot}^{2} \theta } ) = \frac{ 1 }{ { \sin}^{2} \theta - {\sin}^{4} \theta }$

FORMULA:

1 / tan² A = cos² A / sin² A

1 / cot² A = sin² A / cos² A

1 - sin² A = cos² A

sin² A + cos² A = 1

PROOF:

$(1 + \frac{1}{ { \tan}^{2} \theta })(1 + \frac{1}{ { \cot}^{2} \theta } ) = \frac{ 1 }{ { \sin}^{2} \theta - {\sin}^{4} \theta }$

Take L.H.S as

$( \frac{ { \sin}^{2} \theta + {\cos}^{2} \theta}{ { \sin}^{2} } \theta )( \frac{ { \cos}^{2} \theta + {\sin}^{2} \theta}{ { \cos}^{2} } \theta ) \\ \\ \frac{1}{ { \sin}^{2} \theta} \times \frac{1}{ { \cos}^{2} \theta} \\ \\ \frac{1}{{ \sin}^{2} \theta \: { \cos}^{2} \theta }$

Take R.H.S as

$\frac{ 1 }{ { \sin}^{2} \theta - {\sin}^{4} \theta } \\ \\ \frac{1}{ { \sin}^{2} \theta(1 - { \sin}^{2} \theta) } \\ \\ \frac{1}{{ \sin}^{2} \theta \: { \cos}^{2} \theta }$

L. H. S = R. H. S

HENCE PROVED

Answer 2

Answer:

Mark me as brilliant please