Question

Solve the following equations:
(i) $\sin \mathrm{\theta }+\cos \mathrm{\theta }=\sqrt{2}$
(ii) $\sqrt{3}\cos \mathrm{\theta }+\sin \mathrm{\theta }=1$
(iii) $\sin \mathrm{\theta }+\cos \mathrm{\theta }=1$
(iv) $\mathrm{cosec}\mathrm{\theta }=1+\cot \mathrm{\theta }$

Answer 1

(i)
Given:
$\sin \theta +\cos \theta =\sqrt{2}$ ...(i)
The equation is of the form $a\sin \theta +b\cos \theta =c$, where $a=1,b=1$ and $c=\sqrt{2}$.
Let:
$a=r\sin \alpha$and $b=r\cos \alpha$
Now,
$r=\sqrt{a^2+b^2}=\sqrt{1^2+1^2}=\sqrt{2}$ and $\tan \alpha =1\Rightarrow \alpha =\frac{\mathrm{\pi }}{4}$
On putting $a=1=r\sin \alpha$ and $b=1=r\cos \alpha$in equation (i), we get:
$r\sin \alpha \sin \theta +r\cos \alpha \cos \theta =\sqrt{2}$
$$\begin{gathered} \Rightarrow r\cos (\theta -\alpha )=\sqrt{2} \\ \Rightarrow \sqrt{2}\cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=\sqrt{2} \\ \Rightarrow \cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=1 \\ \Rightarrow \cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=\cos 0 \\ \Rightarrow \theta -\frac{\mathrm{\pi }}{4}=n\mathrm{\pi }\pm 0,n\in \mathrm{Z} \\ \Rightarrow \mathrm{\theta }=\mathrm{n\pi }+\frac{\mathrm{\pi }}{4},\mathit{}n\in \mathrm{Z} \\ \Rightarrow \theta =(8n+1)\frac{\mathrm{\pi }}{4},n\in Z \end{gathered}$$

(ii)
Given:
$\sqrt{3}\cos \theta +\sin \theta =1$ ...(ii)
The equation is of the form of $a\cos \theta +b\sin \theta =c$, where $a=\sqrt{3},b=1$ and $c=1$.
Let:
$a=r\cos \alpha$ and $b=r\sin \alpha$
Now,
$r=\sqrt{a^2+b^2}=\sqrt{(\sqrt{3}{)}^2+1^2}=2$ and $\tan \alpha =\frac{b}{a}=\frac{1}{\sqrt{3}}\Rightarrow \alpha =\frac{\mathrm{\pi }}{6}$
On putting $a=\sqrt{3}=r\cos \alpha$ and $b=1=r\sin \alpha$ in equation (ii), we get:
$r\cos \alpha \cos \theta +r\sin \alpha \sin \theta =1$

$$\begin{gathered} \Rightarrow r\cos (\theta -\alpha )\hspace{0.17em}=1 \\ \Rightarrow 2\cos (\theta -\alpha )=1 \\ \Rightarrow \cos \left(\theta -\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2} \\ \Rightarrow \cos \left(\theta -\frac{\mathrm{\pi }}{6}\right)=\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta -\frac{\mathrm{\pi }}{6}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z} \end{gathered}$$

On taking positive sign, we get:
$$\begin{gathered} \theta -\frac{\mathrm{\pi }}{6}=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{3} \\ \Rightarrow \mathrm{\theta }=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6} \\ \Rightarrow \mathrm{\theta }=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z} \\ \Rightarrow \mathrm{\theta }=(4\mathrm{n}+1)\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z} \end{gathered}$$
Now, on taking negative sign of the equation, we get:
$$\begin{gathered} \theta -\frac{\mathrm{\pi }}{6}=2m\mathrm{\pi }-\frac{\mathrm{\pi }}{3},\mathrm{m}\in \mathrm{Z} \\ \Rightarrow \mathrm{\theta }=2\mathrm{m\pi }-\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6},\mathrm{m}\in \mathrm{Z} \\ \Rightarrow \mathrm{\theta }=2\mathrm{m\pi }-\frac{\mathrm{\pi }}{6}=(12\mathrm{m}-1)\frac{\mathrm{\pi }}{6},\mathrm{m}\in \mathrm{Z} \end{gathered}$$

(iii)
Given:
$\sin \theta +\cos \theta =1$ ...(iii)
The equation is of the form $a\sin \theta +b\cos \theta =c$, where $a=1,b=1$ and $c=1$.
Let:
$a=r\sin \alpha$ and $b=r\cos \alpha$
Now,
$r=\sqrt{a^2+b^2}=\sqrt{1^2+1^2}=\sqrt{2}$ and $\tan \alpha =\frac{b}{a}=1\Rightarrow \alpha =\frac{\mathrm{\pi }}{4}$
On putting $a=1=r\sin \alpha$ and $b=1=r\cos \alpha$ in equation (iii), we get:
$r\sin \alpha \sin \theta +r\cos \alpha \cos \theta =1$

$$\begin{gathered} \Rightarrow r\cos (\theta -\alpha )=1 \\ \Rightarrow \sqrt{2}\cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=1 \\ \Rightarrow \cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}} \\ \Rightarrow \cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow \theta -\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4},n\in Z \end{gathered}$$

On taking positive sign, we get:
$$\begin{gathered} \theta -\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{\theta }=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{\theta }=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2},\mathit{}n\in \mathit{}Z \end{gathered}$$
On taking negative sign, we get:
$$\begin{gathered} \theta -\frac{\mathrm{\pi }}{4}=2m\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{\theta }=2m\mathrm{\pi },m\mathit{}\in \mathrm{Z} \end{gathered}$$


(iv)
Given:
$\cos ec\theta =1+cot\theta$

$$\begin{gathered} \Rightarrow \frac{1}{\sin \theta }=1+\frac{\cos \theta }{\sin \theta } \\ \Rightarrow \sin \theta +\cos \theta =1...\left(\mathrm{iv}\right) \end{gathered}$$

The equation is of the form $a\sin \theta +b\cos \theta =c$, where $a=1,b=1$ and $c=1$.
Let:
$a=r\sin \alpha$and $b=r\cos \alpha$
Now,
$r=\sqrt{a^2+b^2}=\sqrt{1^2+1^2}=\sqrt{2}$ and $\tan \alpha =1\Rightarrow \alpha =\frac{\mathrm{\pi }}{4}$
On putting $a=1=r\sin \alpha$ and $b=1=r\cos \alpha$ in equation (iv), we get:

$$\begin{gathered} r\sin \alpha \sin \theta +r\cos \alpha \cos \theta =1 \\ \Rightarrow r\cos (\theta -\alpha )=1 \\ \Rightarrow \sqrt{2}\cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=1 \\ \Rightarrow \cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}} \\ \Rightarrow \cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)=\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow \theta -\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{Z} \end{gathered}$$

On taking positive sign, we get:
$\theta =2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2},n\in Z$

On taking negative sign, we get:
$\theta =2m\pi ,m\in Z$