Question

$\text{Solve the following equations :}\left(i\right)sin\theta +cos\theta =\sqrt{2}\left(ii\right)\sqrt{3}cos\theta +sin\theta =1\left(iii\right)sin\theta +cos\theta =1\left(iv\right)cosec\theta =1+cos\theta \left(v\right)(\sqrt{3}-1)cos\theta +(\sqrt{3}+1)sin\theta =2$

Answer 1

$\left(i\right)sin\theta +cos\theta =\sqrt{2}\text{We have,}sin\theta +cos\theta =\sqrt{2}\Rightarrow \frac{1}{\sqrt{2}}sin\theta +\frac{1}{\sqrt{2}}cos\theta =1\Rightarrow sin\frac{\pi }{4}sin\theta +cos\frac{\pi }{4}cos\theta =1\left[\begin{array}{c}\because cos\frac{\pi }{4}=sin\frac{\pi }{4}=\frac{1}{\sqrt{2}}\end{array}\right]\Rightarrow cos(\theta -\frac{\pi }{4})=cos{0}^{\circ }\Rightarrow \theta -\frac{\pi }{4}=2n\pi ,n\in z\Rightarrow \theta =2n\pi +\frac{\pi }{4},n\in z\therefore \theta =(8n+1)\frac{\pi }{4},n\in z\left(ii\right)\sqrt{3}cos\theta +sin\theta =1\text{Dividing both side by 2,we get}\frac{\sqrt{3}}{2}cos\theta +\frac{1}{2}sin\theta =\frac{1}{2}\Rightarrow cos\frac{\pi }{6}cos\theta +sin\frac{\pi }{6}sin\theta =\frac{1}{2}\left[\begin{array}{c}\because sin=\frac{\pi }{6}=\frac{1}{2},cos\frac{\pi }{6}=\frac{\sqrt{3}}{2}\end{array}\right]\Rightarrow cos(\theta -\frac{\pi }{6})=cos\frac{\pi }{3}\Rightarrow \theta =\frac{\pi }{6}=2n\pm \frac{\pi }{3},n\in z\Rightarrow \theta =2n\pi \pm \frac{\pi }{3}+\frac{\pi }{6},n\in z\Rightarrow \theta =(4n+1)\frac{\pi }{2}Or(12m-1)\frac{\pi }{6},nm\in z\left(iii\right)sin\theta +cos\theta =1\text{We have,}sin\theta +cos\theta =1\text{Divide both side by}\sqrt{2},weget\Rightarrow \frac{1}{\sqrt{2}}sin\theta +\frac{1}{\sqrt{2}}cos\theta =\frac{1}{\sqrt{2}}\Rightarrow sin\frac{\pi }{4}sin\theta +cos\frac{\pi }{4}cos\theta =\frac{1}{\sqrt{2}}\Rightarrow \theta \left(\theta \frac{\pi }{4}\right)=cos\frac{\pi }{4}\Rightarrow \theta =\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4},n\in Z\Rightarrow \theta =2n\pi +\frac{\pi }{2},2n\pi n\in Z\left(iv\right)cosec\theta =1+cos\theta \text{We have,}cosec\theta =1+cos\theta \Rightarrow \frac{1}{sin\theta }=1+\frac{cos\theta }{sin\theta }\Rightarrow 1=sin\theta +cos\theta \text{Divide both side by}\sqrt{2},weget\Rightarrow \frac{1}{\sqrt{2}}sin\theta +\frac{1}{\sqrt{2}}cos\theta =\frac{1}{\sqrt{2}}\Rightarrow sin\frac{\pi }{4}sin\theta +cos\frac{\pi }{4}cos\theta =\frac{1}{\sqrt{2}}\Rightarrow cos(\theta -\frac{\pi }{4})=cos\frac{\pi }{4}\Rightarrow \theta =\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4},n\in Z\therefore \theta (2n\pi +\frac{\pi }{2})Or2n\pi ,n\in Z\left(v\right)(\sqrt{3}-1)cos\theta +(\sqrt{3}+1)sin\theta =2\text{Divide on both sides by}2\sqrt{2}\frac{(\sqrt{3}-1)}{2\sqrt{2}}cos\theta +\frac{(\sqrt{3}+1)}{2\sqrt{2}}sin\theta =\frac{1}{\sqrt{2}}sin(\theta +ta{n}^{-}1\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right))=sin\frac{\pi }{4}\theta =2n\pi +\frac{\pi }{3}Or2n\pi -\frac{\pi }{6}n\in Z$