UESTION 4 1 Given the following expression: (sinx)/(1+cosx)+(cosx)/(sinx) 4.4.1 Prove that: (sinx)/(1+cosx)+(cosx)/(sinx)=(1)/(sinx) 4.4.2 Determine the general solution for which the identity in QUESTION 4.4.1 is undefined. CAPS - Grade 11 If 2cosbeta +1=0 and sinbeta lt 0 determine WITHOUT the use of a calculator and with the aid of a diagram, the value of: 4.1.1 sinbeta 4.1.2 2tan^2beta -cos^2beta 4.2 Simplify (cos(Theta -90^circ )cdot cos(-Theta )cdot tan(360^circ -Theta ))/(sin(180^circ )+Theta ) to a single trigonometric term. 4.3 WITHOUT using a calculator calculate the value of: (2sin510^circ -cos340^circ cdot cos20^circ )/(cos^2)110^(circ ) 4.4

4.4.1 To prove that $\frac {sinx}{1+cosx}+\frac {cosx}{sinx}=\frac {1}{sinx}$, let's start by finding a common denominator for the two fractions on the left side of the equation. The common denominator is $sinx(1+cosx)$. Multiplying the numerator and denominator of each fraction by this common denominator, we get:<br /><br />$\frac {sinx}{1+cosx} \cdot \frac {sinx}{sinx} + \frac {cosx}{sinx} \cdot \frac {1+cosx}{1+cosx} = \frac {sin^2x + cosx(1+cosx)}{sinx(1+cosx)}$<br /><br />Expanding the numerator, we have:<br /><br />$\frac {sin^2x + cosx + cos^2x}{sinx(1+cosx)}$<br /><br />Using the Pythagorean identity $sin^2x + cos^2x = 1$, we can simplify the numerator further:<br /><br />$\frac {1 + cosx}{sinx(1+cosx)}$<br /><br />Canceling out the common factor of $(1+cosx)$ in the numerator and denominator, we are left with:<br /><br />$\frac {1}{sinx}$<br /><br />Therefore, we have proven that $\frac {sinx}{1+cosx}+\frac {cosx}{sinx}=\frac {1}{sinx}$.<br /><br />4.4.2 The identity $\frac {sinx}{1+cosx}+\frac {cosx}{sinx}=\frac {1}{sinx}$ is undefined when the denominator is equal to zero. This occurs when $sinx = 0$. The general solution for $sinx = 0$ is $x = n\pi$, where $n$ is an integer.<br /><br />4.1.1 Given that $2cos\beta +1=0$ and $sin\beta \lt 0$, we can solve for $cos\beta$ by rearranging the equation:<br /><br />$2cos\beta = -1$<br /><br />$cos\beta = -\frac{1}{2}$<br /><br />Since $sin\beta \lt 0$, we know that $\beta$ is in the third or fourth quadrant. In the third quadrant, both sine and cosine are negative, while in the fourth quadrant, only cosine is positive. Therefore, $\beta$ must be in the third quadrant.<br /><br />The reference angle for $cos\beta = -\frac{1}{2}$ is $\frac{\pi}{3}$ or $60^\circ$. Since $\beta$ is in the third quadrant, the general solution for $\beta$ is:<br /><br />$\beta = \pi + \frac{\pi}{3} + 2k\pi = \pi(1 + \frac{1}{3}) + 2k\pi = \frac{4\pi}{3} + 2k\pi$<br /><br />where $k$ is an integer.<br /><br />4.1.2 To find the value of $2tan^{2}\beta - cos^{2}\beta$, we can use the identity $tan\beta = \frac{sin\beta}{cos\beta}$ and the fact that $sin^2\beta + cos^2\beta = 1$.<br /><br />Since $cos\beta = -\frac{1}{2}$, we can find $sin\beta$ using the Pythagorean identity:<br /><br />$sin^2\beta + (-\frac{1}{2})^2 = 1$<br /><br />$sin^2\beta + \frac{1}{4} = 1$<br /><br />$sin^2\beta = 1 - \frac{1}{4}$<br /><br />$sin^2\beta = \frac{3}{4}$<br /><br />Since $\beta$ is in the third quadrant, $sin\beta$ is negative. Therefore, $sin\beta = -\sqrt{\frac{3}{4}} = -\frac{\sqrt{3}}{2}$.<br /><br />Now, we can find $tan\beta$:<br /><br />$tan\beta = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$<br /><br />$tan^2\beta = (\sqrt{3})^2 = 3$<br /><br />$2tan^2\beta = 2 \cdot 3 = 6$<br /><br />Finally, we can find $2tan^2\beta - cos^2\beta$:<br /><br />$2tan^2\beta - cos^2\beta = 6 - (-\frac{1}{2})^2 = 6 - \frac{1}{4} = \frac{23}{4}$<br /><br />4.2 To simplify $\frac {cos(\Theta -90^{\circ })\cdot cos(-\Theta )\cdot tan(360^{\circ }-\Theta