how to Calculate (1╱sin250°＋√3╱cos290°)

Solution Explanation

We want to evaluate the expression

$$E \;=\; \frac{1}{\sin(250^\circ)} \;+\; \frac{\sqrt{3}}{\cos(290^\circ)}.$$

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1. Rewrite the angles in more familiar forms

1. $\sin(250^\circ)$
   Notice $250^\circ = 180^\circ + 70^\circ$.
   We use the identity $\sin(180^\circ + \theta) = -\sin(\theta)$.
   Hence,

   $$\sin(250^\circ) = \sin(180^\circ + 70^\circ) = -\sin(70^\circ).$$

2. $\cos(290^\circ)$
   Notice $290^\circ = 360^\circ - 70^\circ$.
   We use the identity $\cos(360^\circ - \theta) = \cos(\theta)$.
   Hence,

   $$\cos(290^\circ) = \cos(360^\circ - 70^\circ) = \cos(70^\circ).$$

Therefore, the expression becomes

$$E = \frac{1}{-\,\sin(70^\circ)} \;+\; \frac{\sqrt{3}}{\cos(70^\circ)} = -\,\frac{1}{\sin(70^\circ)} \;+\; \frac{\sqrt{3}}{\cos(70^\circ)}.$$

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2. Convert $\sin(70^\circ)$ and $\cos(70^\circ)$

Using the co-function identities:

$$\sin(70^\circ) = \cos(20^\circ), \quad \cos(70^\circ) = \sin(20^\circ).$$

So we can also write

$$E = -\,\frac{1}{\cos(20^\circ)} \;+\; \frac{\sqrt{3}}{\sin(20^\circ)}.$$

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3. Combine into a single fraction

Let’s put everything under one denominator:

$$E = \frac{-\,\sin(20^\circ) + \sqrt{3}\,\cos(20^\circ)}{\sin(20^\circ)\,\cos(20^\circ)}.$$

3.1 Simplify the numerator

Observe the numerator:

$$-\,\sin(20^\circ) + \sqrt{3}\,\cos(20^\circ).$$

A useful identity is that for some $R$ and $\alpha$:

$$\sqrt{3}\,\cos(20^\circ) - \sin(20^\circ) \;=\; R \cos\bigl(20^\circ + \alpha\bigr).$$

By matching coefficients, one can show that $R = 2$ and $\alpha = 30^\circ$, giving

$$\sqrt{3}\,\cos(20^\circ) - \sin(20^\circ) \;=\; 2\,\cos(50^\circ).$$

Hence, the numerator is:

$$-\,\sin(20^\circ) + \sqrt{3}\,\cos(20^\circ) = \sqrt{3}\,\cos(20^\circ) - \sin(20^\circ) = 2\,\cos(50^\circ).$$

3.2 Simplify the denominator

The denominator is $\sin(20^\circ)\,\cos(20^\circ)$. Recall the identity

$$\sin(2\theta) = 2\,\sin(\theta)\cos(\theta).$$

So

$$\sin(20^\circ)\cos(20^\circ) = \tfrac{1}{2}\,\sin(40^\circ).$$

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4. Put it all together

Thus,

$$E = \frac{2\,\cos(50^\circ)}{\sin(20^\circ)\,\cos(20^\circ)} = \frac{2\,\cos(50^\circ)}{\tfrac12\,\sin(40^\circ)} = \frac{2\,\cos(50^\circ)}{1}\,\times \frac{2}{\sin(40^\circ)} = 4\,\frac{\cos(50^\circ)}{\sin(40^\circ)}.$$

But notice $\cos(50^\circ) = \sin(40^\circ)$ because $\cos(50^\circ) = \sin(90^\circ - 50^\circ) = \sin(40^\circ)$.

Hence,

$$E = 4 \times \frac{\sin(40^\circ)}{\sin(40^\circ)} = 4.$$

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5. Final result

$$\boxed{4}.$$

Therefore,

$$\frac{1}{\sin(250^\circ)} + \frac{\sqrt{3}}{\cos(290^\circ)} = 4.$$