Question

If $\sin \theta +\cos \theta =\sqrt{2}$, then find the value of $\tan \theta +\cot \theta$.

Answer 1

Consider the given equation,

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

Taking square both sides,

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

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

$2\sin \theta \cos \theta =1$

$\sin \theta \cos \theta =\frac{1}{2}$ ……(2)

Now, divided by $\cos \theta$ in equation 1^st , we get

$\tan \theta +1=\frac{\sqrt{2}}{\cos \theta }$ ….(3)

Again divided by $\sin \theta$ in equation 1^st, we get

$1+\cot \theta =\frac{\sqrt{2}}{\sin \theta }$ ….(4)

Add equation 1^st and 2^nd , we get

$\tan \theta +\cot \theta +2=\frac{\sqrt{2}}{\cos \theta }+\frac{\sqrt{2}}{\sin \theta }$

$\tan \theta +\cot \theta =\sqrt{2}.\left(\frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }\right)-2$ ……(5)

Now, from equation 1^st ,2^nd and 5^th ,we get

$\tan \theta +\cot \theta =\sqrt{2}.\left(\frac{\sqrt{2}}{\frac{1}{2}}\right)-2$

$\tan \theta +\cot \theta =2$

Hence, this is the answer.