Question

[1+1/tan^x][1+1/cot^x]=1/sin^x-sin^x​

Answer 1

Correct Statement :-

$\sf \: Prove\:that\:\bigg(1 + \dfrac{1}{ {tan}^{2} x}\bigg)\bigg(1 + \dfrac{1}{ {cot}^{2}x}\bigg) = \dfrac{1}{ {sin}^{2}x - {sin}^{4}x}$

Identities Used :-

$\boxed{\bf \: {sin}^{2}x + {cos}^{2}x = 1}$

$\boxed{\bf \:{cos}^{2}x = 1 - {sin}^{2}x}$

$\boxed{ \bf \:tanx = \dfrac{sinx}{cosx}}$

$\boxed{ \bf \:cotx = \dfrac{cosx}{sinx}}$

Solution :-

Consider,

$\sf \:\bigg(1 + \dfrac{1}{ {tan}^{2} x}\bigg)\bigg(1 + \dfrac{1}{ {cot}^{2}x}\bigg)$

$\sf \: = \: \: \bigg(1 + \dfrac{ {cos}^{2}x }{ {sin}^{2} x}\bigg)\bigg(1 + \dfrac{ {sin}^{2}x }{ {cos}^{2}x}\bigg)$

$\sf \: = \: \: \bigg(\dfrac{{sin}^{2}x + {cos}^{2}x}{ {sin}^{2}x}\bigg)\bigg(\dfrac{{cos}^{2}x+{sin}^{2}x }{ {cos}^{2}x}\bigg)$

$\sf \: = \: \: \dfrac{1}{{sin}^{2}x} \times\dfrac{1}{{cos}^{2}x}$

$\sf \: = \: \: \dfrac{1}{{sin}^{2}x {cos}^{2} x}$

$\sf \: = \: \: \dfrac{1}{{sin}^{2}x (1 - {sin}^{2} x)}$

$\sf \: = \: \: \dfrac{1}{{sin}^{2}x - {sin}^{4} x}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

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

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

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

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

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

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

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

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

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

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

${ \green{ \bf{ sec θ = 1/cos θ }}}$

${ \green{ \bf{ cot θ = 1/tan θ }}}$

${ \green{ \bf{sin θ = 1/cosec θ }}}$

${ \green{ \bf{ cos θ = 1/sec θ }}}$

${ \green{ \bf{ tan θ = 1/cot θ}}}$

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

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

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

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

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

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

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

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

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

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

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