Question

Solve:
$(\sin \theta +\cos \theta )(\tan \theta +\cot \theta )=\csc \theta +\cot \theta$

Answer 1

We have,

$(\sin \theta +\cos \theta )(\tan \theta +\cot \theta )=\csc \theta +\cot \theta$

$\Rightarrow (\sin \theta +\cos \theta )(\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta })=\csc \theta +\cot \theta$

$\Rightarrow (\sin \theta +\cos \theta )\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }\right)=\csc \theta +\cot \theta$

$\Rightarrow (\sin \theta +\cos \theta )\left(\frac{1}{\sin \theta \cos \theta }\right)=\csc \theta +\cot \theta$

$\Rightarrow \frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }=\csc \theta +\cot \theta$

$\Rightarrow \frac{\sin \theta }{\sin \theta \cos \theta }+\frac{\cos \theta }{\sin \theta \cos \theta }=\csc \theta +\cot \theta$

$\Rightarrow \sec \theta +\csc \theta =\csc \theta +\cot \theta$

$\Rightarrow \sec \theta =\cot \theta$

$\Rightarrow \frac{1}{\cos \theta }=\frac{\cos \theta }{\sin \theta }$

$\Rightarrow {\cos }^2\theta =\sin \theta$

$\Rightarrow 1-{\sin }^2\theta ={\sin }^2\theta$

$\Rightarrow 1=2{\sin }^2\theta$

$\Rightarrow \frac{1}{2}={\sin }^2\theta$

$\Rightarrow {\sin }^2\theta =\frac{1}{2}$

$\Rightarrow {\sin }^2\theta ={\left(\frac{1}{\sqrt{2}}\right)}^2$

On comparing both side and we get,

${\sin }^2\theta ={\sin }^2\frac{\pi }{4}$

$\theta =\frac{\pi }{4}\left(\text{Principal}\text{value}\right)$

Hence, this is the answer.