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# Prove that: $(\sin A+\cos A)(1-\sin A\cos A)={{\sin }^{3}}A+{{\cos }^{3}}A$  Hint: We will begin with the left hand side of the equation (1) and multiply all the terms by opening the brackets. Then we will get ${{\sin }^{2}}A$ and ${{\cos }^{2}}A$ as two among all the terms and then we will convert ${{\sin }^{2}}A$ in terms of ${{\cos }^{2}}A$ and ${{\cos }^{2}}A$ in terms of ${{\sin }^{2}}A$ using the trigonometry identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. And after doing all these we will see every other term gets cancelled with each other leaving only those terms which are there in the right hand side of equation (1).

It is mentioned in the question that $(\sin A+\cos A)(1-\sin A\cos A)={{\sin }^{3}}A+{{\cos }^{3}}A......(1)$

Now beginning with the left hand side of the equation (1) we get,

$\Rightarrow (\sin A+\cos A)(1-\sin A\cos A).........(2)$

Now opening the brackets and multiplying both the terms in equation (2) we get,

\begin{align} & \Rightarrow \sin A(1-\sin A\cos A)+\cos A(1-\sin A\cos A) \\ & \Rightarrow \sin A-{{\sin }^{2}}A\cos A+\cos A-\sin A{{\cos }^{2}}A..........(3) \\ \end{align}

Now converting ${{\sin }^{2}}A$ in terms of ${{\cos }^{2}}A$ and ${{\cos }^{2}}A$ in terms of ${{\sin }^{2}}A$ using the trigonometry identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ we get,

$\Rightarrow \sin A-(1-{{\cos }^{2}}A)\cos A+\cos A-\sin A(1-{{\sin }^{2}}A)..........(4)$

Now again opening the brackets and multiplying the terms in equation (4) we get,

$\Rightarrow \sin A-\cos A+{{\cos }^{3}}A+\cos A-\sin A+{{\sin }^{3}}A..........(5)$

Now cancelling the similar terms in equation (5) we get,

$\Rightarrow {{\cos }^{3}}A+{{\sin }^{3}}A..........(6)$

Hence from equation (6) we can say that the left hand side is equal to the right hand side in equation (1). Hence we have proved the given expression.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (3) but here the key is to convert ${{\sin }^{2}}A$ in terms of ${{\cos }^{2}}A$ and ${{\cos }^{2}}A$ in terms of ${{\sin }^{2}}A$. Also to make things clearer we have done grouping which will directly help us to see the similar terms getting cancelled.
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