Question

cos^2x/1- tan x + sin^3/sin x - cos x = 1+ sin x cos x​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

Prove that

$\rm :\longmapsto\:\dfrac{ {cos}^{2} x}{1 - tanx} + \dfrac{ {sin}^{3} x}{sinx - cosx} = 1 + sinxcosx$

Identities Used :-

$\red{ \boxed{ \sf{tanx \: = \: \frac{sinx}{cosx} }}}$

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

$\red{ \boxed{ \sf{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) }}}$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\:\dfrac{ {cos}^{2} x}{1 - tanx} + \dfrac{ {sin}^{3} x}{sinx - cosx}$

$\rm \: \: = \: \:\dfrac{ {cos}^{2} x}{1 - tanx} + \dfrac{ {sin}^{3} x}{sinx - cosx}$

$\rm \: \: = \: \:\dfrac{ {cos}^{2} x}{1 - \dfrac{sinx}{cosx} } + \dfrac{ {sin}^{3} x}{sinx - cosx}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \because \: \: tanx = \frac{sinx}{cosx} }}}$

$\rm \: \: = \: \:\dfrac{ {cos}^{2} x}{ \: \: \: \dfrac{cosx - sinx }{cosx} \: \: \: } + \dfrac{ {sin}^{3} x}{sinx - cosx}$

$\rm \: \: = \: \:\dfrac{ {cos}^{3} x}{ \: \: \: cosx - sinx \: \: \: } + \dfrac{ {sin}^{3} x}{sinx - cosx}$

$\rm \: \: = \: \:\dfrac{ {cos}^{3} x}{ \: \: \: cosx - sinx \: \: \: } - \: \dfrac{ {sin}^{3} x}{cosx - sinx}$

$\rm \: \: = \: \:\dfrac{ {cos}^{3} x - {sin}^{3} x}{ \: \: \: cosx - sinx \: \: \: }$

$\: \: \: \: \: \: \: \red{ \boxed{ \sf{ \because \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}}}$

$\rm \: \: = \: \:\dfrac{ \cancel{ (cosx - sinx)} \: \: ({cos}^{2} x + cosx \: sinx + {sin}^{2}x)}{ \: \: \: \cancel{ cosx - sinx \:} \: \: }$

$\rm \: \: = \: {sin}^{2}x + {cos}^{2}x + sinx \: cosx$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \because \: {sin}^{2}x + {cos}^{2}x = 1 }}}$

$\rm \: \: = \: 1 + sinx \: cosx$

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