Question

If alpha n beta are angles in the first quadrant tan alpha=1/7,sin beta=1/√10 then find value of (alpha +2beta)

Answer 1

I hope this will help u

Answer 2

Answer:

$\alpha + 2\beta = 45\degree$

Step-by-step explanation:

$Given \: \alpha \: and \: \beta \\ are \: angles \: in \: the \\ first \: quadrant$

$tan\alpha = \frac{1}{7}\:---(1)$

$sin\beta = \frac{1}{\sqrt{10}}\:---(2)$

$cos\beta\\ = \sqrt{1-sin^{2}\beta}\\=\sqrt{1-\left(\frac{1}{\sqrt{10}}\right)^{2}}$

$=\sqrt{1-\frac{1}{10}}\\=\sqrt{\frac{10-1}{10}}\\=\sqrt{\frac{9}{10}}\\=\frac{3}{\sqrt{10}}\:--(3)$

$tan\beta \\= \frac{sin\beta}{cos\beta}\\=\frac{1}{3}\:---(4)$

$tan2\beta = \frac{2tan\beta}{1-tan^{2}\beta}\\=\frac{2\times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\\=\frac{3}{4}\:---(5)$

$Now,\\tan(\alpha+2\beta)\\=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times \frac{3}{4}}\\=\frac{\frac{4+21}{28}}{\frac{28-3}{28}}\\=\frac{25}{25}\\=1\\=tan45\degree$

$\implies \alpha + 2\beta = 45\degree$

Therefore,

$\alpha + 2\beta = 45\degree$

•••♪