If$\ an{\alpha = \frac{1}{\sqrt{x\left( x^{2} + x + 1 ight)}}}, an\beta = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}\$and$\ an\gamma = \sqrt{x^{- 3} + x^{- 2} + x^{- 1},}$ then$\ \alpha + \beta\$ is

Question

If$\ 	an{\alpha = \frac{1}{\sqrt{x\left( x^{2} + x + 1 ight)}}},	an\beta = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}\$and$\ 	an\gamma = \sqrt{x^{- 3} + x^{- 2} + x^{- 1},}$ then$\ \alpha + \beta\$ is

Dr. Abubacker · Accepted Answer

(a)
We have, $	an{(\alpha + \beta) = \frac{	an{\alpha + 	an\beta}}{1 - 	an{\alpha	an\beta}}}$
$= \frac{\frac{1}{\sqrt{x(x^{2} + x + 1)}} + \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}}{1 - \frac{1}{\sqrt{x(x^{2} + x + 1)}}.\frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}}$
$= \frac{(1 + x)\sqrt{x^{2} + x + 1}}{\sqrt{x}.x(x + 1)}$
$= \sqrt{x^{- 3} + x^{- 2} + x^{- 1}} = 	an{\gamma\ (given)}$
$	herefore\ \alpha + \beta = \gamma$

Q. If$\ \tan{\alpha = \frac{1}{\sqrt{x\left( x^{2} + x + 1 \right)}}},\tan\beta = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}\$and$\ \tan\gamma = \sqrt{x^{- 3} + x^{- 2} + x^{- 1},}$ then$\ \alpha + \beta\$ is

Explanation

Reviewed by Dr. Abubacker

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