Question

sin² x
1 + cot x
cos? *
1 + tan x
= sin x cos x​

Answer 1

$lets \: Consider, \\ \sf \:\bigg(1 + \dfrac{1}{ {tan}^{2} x}\bigg)\bigg(1 + \dfrac{1}{ {cot}^{2}x}\bigg) \\ \sf \: = \: \: \bigg(1 + \dfrac{ {cos}^{2}x }{ {sin}^{2} x}\bigg)\bigg(1 + \dfrac{ {sin}^{2}x }{ {cos}^{2}x}\bigg)\\ \\ \sf \: = \: \: \bigg(\dfrac{{sin}^{2}x + {cos}^{2}x}{ {sin}^{2}x}\bigg)\bigg(\dfrac{{cos}^{2}x+{sin}^{2}x }{ {cos}^{2}x}\bigg)\\ \\ \sf \: = \: \: \dfrac{1}{{sin}^{2}x} \times\dfrac{1}{{cos}^{2}x}\\ \\ \sf \: = \: \: \dfrac{1}{{sin}^{2}x {cos}^{2} x} \\ \sf \: = \: \: \dfrac{1}{{sin}^{2}x (1 - {sin}^{2} x)}\\ \sf \: = \: \: \dfrac{1}{{sin}^{2}x - {sin}^{4} x}$

${ \red{ \bf{ Information \: related \: to \:Trigonometry:}}}$

${ \green{ \bf{ sin θ = Opposite Side/Hypotenuse }}}$

${ \green{ \bf{ cos θ = Adjacent Side/Hypotenuse }}}$

${ \green{ \bf{tan θ = Opposite Side/Adjacent Side }}}$

${ \green{ \bf{sec θ = Hypotenuse/Adjacent Side }}}$

${ \green{ \bf{ cosec θ = Hypotenuse/Opposite Side }}}$

${ \green{ \bf{ cot θ = Adjacent Side/Opposite Side }}}$

${ \red{ \bf{Their \: reciprocal \: Identities: }}}$

${ \green{ \bf{ cosec θ = 1/sin θ }}}$

${ \green{ \bf{ sec θ = 1/cos θ }}}$

${ \green{ \bf{ cot θ = 1/tan θ }}}$

${ \green{ \bf{sin θ = 1/cosec θ }}}$

${ \green{ \bf{ cos θ = 1/sec θ }}}$

${ \green{ \bf{ tan θ = 1/cot θ}}}$

${ \red{ \bf{ Their \: co-function \: Identities: }}}$

${ \green{ \bf{ sin (90°−x) = cos x }}}$

${ \green{ \bf{cos (90°−x) = sin x }}}$

${ \green{ \bf{ tan (90°−x) = cot x }}}$

${ \green{ \bf{ cot (90°−x) = tan x }}}$

${ \green{ \bf{ sec (90°−x) = cosec x }}}$

${ \green{ \bf{ cosec (90°−x) = sec x }}}$

${ \red{ \bf{ Their \: fundamental \: trigonometric \: identities: }}}$

${ \green{ \bf{ sin²θ + cos²θ = 1 }}}$

${ \green{ \bf{ sec²θ - tan²θ = 1 }}}$

${ \green{ \bf{ cosec²θ - cot²θ = 1 }}}$