Question

if tan x=a/b, prove that asin2x+bcos2x=b

Answer 1

$\textbf{Given:}$

$tan\,x=\frac{a}{b}$

$\text{We know that,}$

$\implies\boxed{\begin{minipage}{3cm} \bf\,sin\,2A=\frac{2\,tanA}{1+tan^2A}\\\\cos\,2A=\frac{1-tan^2A}{1+tan^2A} \end{minipage}}$

$\text{Now,}$

$a\,sin2x+b\,cos2x$

$=\displaystyle\,a[\frac{2\,tanx}{1+tan^2x}]+b[\frac{1-tan^2x}{1+tan^2x}]$

$=\displaystyle\,a[\frac{2(\frac{a}{b})}{1+\frac{a^2}{b^2}}]+b[\frac{1-\frac{a^2}{b^2}}{1+\frac{a^2}{b^2}}]$

$=\displaystyle\,[\frac{\frac{2a^2}{b}}{\frac{a^2+b^2}{b^2}}]+b[\frac{\frac{b^2-a^2}{b^2}}{\frac{a^2+b^2}{b^2}}]$

$=\displaystyle\,\frac{\frac{2a^2b^2}{b}}{a^2+b^2}+b[\frac{b^2-a^2}{a^2+b^2}]$

$=\displaystyle\,\frac{2a^2b}{a^2+b^2}+\frac{b^3-a^2b}{a^2+b^2}$

$=\displaystyle\,\frac{2a^2b+b^3-a^2b}{a^2+b^2}$

$=\displaystyle\,\frac{a^2b+b^3}{a^2+b^2}$

$=\displaystyle\,\frac{b(a^2+b^2)}{a^2+b^2}$

$=b$

$\implies\boxed{\bf\,a\,sin2x+b\,cos2x=b}$

Answer 2

I hope this answer helps you.