Question
Prove the following trigonometric identities.
If $\sin\theta+2\cos\theta=1$ prove that $2\sin\theta-\cos\theta=2.$
If $\sin\theta+2\cos\theta=1$ prove that $2\sin\theta-\cos\theta=2.$
✓
Answer
If is given that,
$\text{L.H.S}=\sin\theta+2\cos\theta=1$
$\Rightarrow\ 2\cos\theta=1-\sin\theta\ .....(\text{i})$
Squarign both sides, we get
$\Rightarrow\ (2\cos\theta)^2=(1-\sin\theta)^2$
$\Rightarrow\ 4\cos^2\theta=1+\sin^2\theta-2\sin\theta$
$\Rightarrow\ 4\cos^2\theta=1+\sin^2\theta-2\sin\theta-1+1$
$\Rightarrow\ 4\cos^2\theta=2-2\sin\theta-(1-\sin^2\theta)$
$\Rightarrow\ 4\cos^2\theta=2-2\sin\theta-\cos^2\theta$
$\Rightarrow\ 5\cos^2\theta=2(1-\sin\theta)$
$\Rightarrow\ 5\cos^2\theta=4\cos\theta \ [\text{Using (i)}]$
$\Rightarrow\ \cos\theta(5\cos\theta-4)=0$
$\Rightarrow\ \cos\theta=0,\cos\theta=\frac{4}{5}$
Putting $\cos\theta=\frac{4}{5}\text{ in }\sin\theta+2\cos\theta=1,$ we get
$\Rightarrow\ \sin\theta=1-2\cos\theta=1-2\Big(\frac{4}{5}\Big)=-\frac{3}{5}$
This is not possible.
Putting $\cos\theta=0\text{ in }\sin\theta+2\cos\theta=1,$ we get
$\sin\theta=1$
Thus, the value of $2\sin\theta-\cos\theta=2(1)-0=2.=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$\text{L.H.S}=\sin\theta+2\cos\theta=1$
$\Rightarrow\ 2\cos\theta=1-\sin\theta\ .....(\text{i})$
Squarign both sides, we get
$\Rightarrow\ (2\cos\theta)^2=(1-\sin\theta)^2$
$\Rightarrow\ 4\cos^2\theta=1+\sin^2\theta-2\sin\theta$
$\Rightarrow\ 4\cos^2\theta=1+\sin^2\theta-2\sin\theta-1+1$
$\Rightarrow\ 4\cos^2\theta=2-2\sin\theta-(1-\sin^2\theta)$
$\Rightarrow\ 4\cos^2\theta=2-2\sin\theta-\cos^2\theta$
$\Rightarrow\ 5\cos^2\theta=2(1-\sin\theta)$
$\Rightarrow\ 5\cos^2\theta=4\cos\theta \ [\text{Using (i)}]$
$\Rightarrow\ \cos\theta(5\cos\theta-4)=0$
$\Rightarrow\ \cos\theta=0,\cos\theta=\frac{4}{5}$
Putting $\cos\theta=\frac{4}{5}\text{ in }\sin\theta+2\cos\theta=1,$ we get
$\Rightarrow\ \sin\theta=1-2\cos\theta=1-2\Big(\frac{4}{5}\Big)=-\frac{3}{5}$
This is not possible.
Putting $\cos\theta=0\text{ in }\sin\theta+2\cos\theta=1,$ we get
$\sin\theta=1$
Thus, the value of $2\sin\theta-\cos\theta=2(1)-0=2.=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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