Question

Hello, can you please help me solve question number 8 !

Answer 1

1

Step-by-step explanation:

`8)\text{ }\sin A\text{ cosA tanA + }\frac{2\sin Acos^3A}{\sin\text{ 2A}}`
`\begin{gathered} \sin A\text{ cosA tanA + }\frac{2\sin Acos^3A}{\sin\text{ 2A}} \\ \tan \text{ A = }(\sin A)/(\cos A) \\ \sin A\text{ cosA }*(\sin A)/(\cos A)\text{ + }\frac{2\sin Acos^3A}{\sin\text{ 2A}} \\ =\text{ sinA(sinA) + }\frac{2\sin Acos^3A}{\sin\text{ 2A}} \end{gathered}`
`\begin{gathered} =sin^2A\text{ + }\frac{2\sin Acos^3A}{\sin\text{ 2A}} \\ =\text{ }\frac{\sin ^2A(sin2A)+2sinAcos^3A}{\sin \text{ 2A}} \\ \sin 2A\text{ = 2sinAcosA} \\ =\text{ }\frac{\sin ^2A(sin2A)+2sinAcos^3A}{2\sin \text{ AcosA}} \\ =\text{ }\frac{\sin ^2A(2sinA\cos A)+2sinAcos^3A}{2\sin \text{ AcosA}} \end{gathered}`
`\begin{gathered} =\text{ }\frac{\sin^2A(2sinA\cos A)+2sinAcos^3A}{2\sin\text{ AcosA}} \\ =\frac{2\sin ^3A(\cos A)+2sinAcos^3A}{2\sin \text{ AcosA}} \\ =\frac{2\sin ^3A\cos A+2sinAcos^3A}{2\sin \text{ AcosA}} \\ =\text{ }\frac{\cos A(2\sin ^3A)+\cos A(2sinAcos^2A)}{2\sin \text{ AcosA}} \end{gathered}`
`\begin{gathered} =\text{ }\frac{\cos A(2\sin^3A)+\cos A(2sinAcos^2A)}{2\sin\text{ AcosA}} \\ =\frac{\cos A\lbrack(2\sin^3A)+(2sinAcos^2A)\rbrack}{\text{cosA(}2\sin A)} \\ =((2\sin^3A)+(2sinAcos^2A))/(2\sin A) \\ =(2\sin A(\sin ^2A+cos^2A))/(2\sin A) \end{gathered}`
`\begin{gathered} =sin^2A+cos^2A \\ \text{reca ll:} \\ \text{ sin}^2x+cos^2x\text{ = 1} \\ \\ \text{Hence, }sin^2A+cos^2A\text{ = 1} \end{gathered}`