Question

cos square theta by 1 minus tan theta + sin cube theta by sin theta minus cos theta is equal to 1 + sin theta cos theta​

Answer 1

Step-by-step explanation:

We have,

$\rm \dfrac{\cos ^2 \theta}{1- \tan \theta} + \dfrac{\sin ^3 \theta}{\sin \theta - \cos \theta} = 1+ \sin \theta \cos \theta$

LHS,

$\rm \dfrac{\cos ^2 \theta}{1- \tan \theta} + \dfrac{\sin ^3 \theta}{\sin \theta - \cos \theta}$

$\rm \dfrac{\cos ^2 \theta}{1- \dfrac{\sin \theta}{\cos \theta}} + \dfrac{\sin ^3 \theta}{\sin \theta - \cos \theta}$

$\rm \dfrac{\cos ^2 \theta}{\dfrac{\cos \theta -\sin \theta}{\cos \theta}} + \dfrac{\sin ^3 \theta}{\sin \theta - \cos \theta}$

$\rm \dfrac{\cos ^3 \theta}{\cos \theta - \sin \theta} + \dfrac{\sin ^3 \theta}{\sin \theta - \cos \theta}$

$\rm \dfrac{\cos ^3 \theta}{\cos \theta - \sin \theta} - \dfrac{\sin ^3 \theta}{\cos \theta - \sin \theta}$

$\rm \dfrac{\cos ^3 \theta ( \cos \theta - \sin \theta ) - \sin ^3 \theta (\cos \theta - \sin \theta )}{(\cos \theta - \sin \theta)(\cos \theta - \sin \theta)}$

$\rm \dfrac{\cos ^3 \theta -\sin ^3 \theta}{\cos θ - \sin θ}$

$\rm \dfrac{(\cos \theta - \sin \theta ) (\cos ^2 \theta -\sin ^2 \theta + \sin \theta \cos \theta)}{\cos θ - \sin θ}$

$\rm = 1 + \sin \theta \cos \theta$

Answer 2

Question

Prove that

$\rm :\longmapsto\:\dfrac{ {cos}^{2}\theta }{1 - tan\theta } + \dfrac{ {sin}^{3}\theta }{sin\theta - cos\theta } = 1 + sin\theta cos\theta$

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

Consider, LHS

$\rm :\longmapsto\:\dfrac{ {cos}^{2}\theta }{1 - tan\theta } + \dfrac{ {sin}^{3}\theta }{sin\theta - cos\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ tan\theta = \frac{sin\theta }{cos\theta }}}}$

So, using this identity, we get

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

$\rm \: = \: \dfrac{ {cos}^{2}\theta }{ \dfrac{cos\theta - sin\theta }{cos\theta } } + \dfrac{ {sin}^{3}\theta }{sin\theta - cos\theta }$

$\rm \: = \: \dfrac{ {cos}^{3} \theta }{cos\theta - sin\theta } + \dfrac{ {sin}^{3}\theta }{sin\theta - cos\theta }$

$\rm \: = \: \dfrac{ {cos}^{3} \theta }{cos\theta - sin\theta } + \dfrac{ {sin}^{3}\theta }{ - (cos\theta - sin\theta) }$

$\rm \: = \: \dfrac{ {cos}^{3} \theta }{cos\theta - sin\theta } - \dfrac{ {sin}^{3}\theta }{cos\theta - sin\theta}$

$\rm \: = \: \dfrac{ {cos}^{3} \theta - {sin}^{3} \theta }{cos\theta - sin\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) \: }}}$

So, using this identity, we get

$\rm \: = \: \dfrac{ \cancel{(cos\theta - sin\theta )} \: ( {cos}^{2} \theta + {sin}^{2}\theta + cos\theta sincos) }{ \cancel{cos\theta - sin\theta }}$

$\rm \: = \: {cos}^{2}\theta + {sin}^{2}\theta + cos\theta sin\theta$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {cos}^{2}x + {sin}^{2}x = 1}}}$

So, using this identity, we get

$\rm \: = \: 1 + sin\theta \: cos\theta$

Hence,

$\red{\rm\implies \:\boxed{\sf{ \dfrac{ {cos}^{2}\theta }{1 - tan\theta } + \dfrac{ {sin}^{3}\theta }{sin\theta - cos\theta } = 1 + sin\theta cos\theta }}}$

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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