Question

if cosec alpha +cot alpha =2√3than prove that cos alpha =2\√5

Answer 1

Answer on Question #80411 – Math – Trigonometry

Question

$$\begin{array}{l} \text{Cosec } (\alpha) + \cot(\alpha) = 2\sqrt{3} \\ \text{Check whether } \cos(\alpha) = 2/\sqrt{5} \end{array}$$

Solution

$$\begin{array}{l} 1 / \sin(\alpha) + \cos(\alpha) / \sin(\alpha) = 2^*\sqrt{3} \quad \sin(\alpha) \neq 0 \\ (1 + \cos(\alpha)) / \sin(\alpha) = 2^*\sqrt{3} \\ ((1 + \cos(\alpha)) / \sin(\alpha))^2 = (2^*\sqrt{3})^2 \\ (1 + 2 \cos(\alpha) + \cos^2(\alpha)) / \sin^2(\alpha) = 12 \\ 1 + 2 \cos(\alpha) + \cos^2(\alpha) = 12 (1 - \cos^2(\alpha)) \\ 13 \cos^2(\alpha) + 2 \cos(\alpha) - 11 = 0 \quad \cos(\alpha) = x \\ 13x^2 + 2x - 11 = 0 \\ X_{1,2} = (-2 \pm \sqrt{(2^2 - 4*13*(-11))}/2*13 \quad x_1 = -1 \quad x_2 = 11/13 \\ \end{array}$$

If $\cos(\alpha) = -1$, then in this case $\sin(\alpha) = 0$, hence cosec $(\alpha)$ and $\cot(\alpha)$ do not exist, therefore $\cos(\alpha) = 11/13$.

**Answer**: $\cos(\alpha) = 11/13$

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