If $$an \alpha = \frac{a}{b}$$ then the value of $$\sin \alpha + \cos \alpha$$ is:

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RRB NTPC Exam 27 Jan 2021 click here to view all questions

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. Akhilesh ? answered Apr 1 '25 at 20:40

undefined Akhilesh ? answered Apr 3 '25 at 21:00

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