If $$ an \alpha = \frac{a}{b} $$ then the value of $$ \sin \alpha + \cos \alpha $$ is:
If $$ an \alpha = \frac{a}{b} $$ then the value of $$ \sin \alpha + \cos \alpha $$ is:
A$$ \frac{a+b}{\sqrt{a^2+b^2}} $$
B$$ \frac{\sqrt{a^2+b^2}}{a+b} $$
C$$ \frac{a+b}{\sqrt{a^2-b^2}} $$
D$$ \frac{\sqrt{a^2-b^2}}{a+b} $$
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correct answer is: option a : \(\frac{a+b}{\sqrt{a^2+b^2}}\)
Explanation: Given \(\tan \alpha = \frac{a}{b}\), we can consider a right-angled triangle where the opposite side is *a* and the adjacent side is *b*. Then, the hypotenuse will be \(\sqrt{a^2 + b^2}\). Therefore, \(\sin \alpha = \frac{a}{\sqrt{a^2 + b^2}}\) and \(\cos \alpha = \frac{b}{\sqrt{a^2 + b^2}}\). So, \(\sin \alpha + \cos \alpha = \frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{a^2 + b^2}} = \frac{a+b}{\sqrt{a^2 + b^2}}\).
- Option a: \(\frac{a+b}{\sqrt{a^2+b^2}}\) This is the correct result obtained from the trigonometric relationships and the Pythagorean theorem.
- Option b: \(\frac{\sqrt{a^2+b^2}}{a+b}\) This is the reciprocal of the correct answer.
- Option c: \(\frac{a+b}{\sqrt{a^2-b^2}}\) This option contains \(\sqrt{a^2-b^2}\), which is incorrect based on the given information about \(\tan \alpha\) and the Pythagorean theorem. It would be applicable if we had to deal with secant and tangent where \(\sec^2{\alpha} - \tan^2{\alpha} = 1\).
- Option d: \(\frac{\sqrt{a^2-b^2}}{a+b}\) This option also contains \(\sqrt{a^2-b^2}\), which is incorrect based on the given right triangle setup.
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