Question

prove that cos^2A/1-tanA+sin^3A/sinA-cosA=1+sinAcosA please!!!!

Answer 1

l hope that it will help you

Answer 2

Answer:  Proved.

Step-by-step explanation:  We are given to prove the following:

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

To prove this, we need the following identities:

$(i)~\dfrac{\sin\theta}{\cos\theta}=\tan\theta,\\\\(ii)~\cos^2\theta+\sin^2\theta=1,\\\\(iii)~a^3-b^3=(a-b)(a^2+b^2+ab).$

We have

$L.H.S.\\\\=\dfrac{\cos^2\theta}{1-\tan\theta}+\dfrac{\sin^3\theta}{\sin\theta-\cos\theta}\\\\\\=\dfrac{\cos^2\theta}{1-\frac{\sin\theta}{\cos\theta}}+\dfrac{\sin^3\theta}{\sin\theta-\cos\theta}\\\\\\=\dfrac{\cos^2\theta}{\frac{\cos\theta-\sin\theta}{\cos\theta}}}+\dfrac{\sin^3\theta}{\sin\theta-\cos\theta}\\\\\\=\dfrac{\cos^3\theta}{\cos\theta-\sin\theta}-\dfrac{\sin^3\theta}{\cos\theta-\sin\theta}\\\\\\=\dfrac{\cos^3\theta-\sin^3\theta}{\cos\theta-\sin\theta}\\\\\\=\dfrac{(\cos\theta-\sin\theta)(\cos^2\theta+\sin^2\theta+\cos\theta\sin\theta)}{(\cos\theta-\sin\theta)}\\\\\\=\cos^2\theta+\sin^2\theta+\sin\theta\cos\theta\\\\\\=1+\sin\theta\cos\theta\\\\=R.H.S.$

Hence proved.