Question

Question 5
If $sin\theta +2cos\theta =1$, then prove that $2sin\theta -cos\theta =2$.

Answer 1

Given, $sin\theta +2cos\theta =1$

On squaring both sides, we get;

$(sin\theta +2cos\theta {)}^2=1$

$\Rightarrow si{n}^2\theta +4co{s}^2\theta +4sin\theta .cos\theta =1$

$\Rightarrow (1-co{s}^2\theta )+4(1-si{n}^2\theta )+4sin\theta .cos\theta =1$

$[\because si{n}^2\theta +co{s}^2\theta =1]$

$\Rightarrow -co{s}^2\theta -4si{n}^2\theta +4sin\theta .cos\theta =-4$

$\Rightarrow 4si{n}^2\theta +co{s}^2\theta -4sin\theta .cos\theta =4$

$\Rightarrow (2sin\theta -cos\theta {)}^2=4$

$[\because a^2+b^2-2ab=(a-b{)}^2]$

$\Rightarrow 2sin\theta -cos\theta =2$