If $ \sin \theta+\cos \theta=\sqrt{2} \cos \left(90^{\circ}-\theta\right) $, find $ \cot \theta $.

Given:

\( \sin \theta+\cos \theta=\sqrt{2} \cos \left(90^{\circ}-\theta\right) \)

To do:

We have to find \( \cot \theta \).

Solution:  

$\sin \theta+\cos \theta=\sqrt{2} \cos (90^{\circ}-\theta)$

$\Rightarrow \sin \theta+\cos \theta=\sqrt{2} \sin \theta$        [Since $\cos (90^{\circ}-\theta)=\sin \theta$]

$\Rightarrow \cos \theta=\sqrt{2} \sin \theta-\sin \theta$

$\Rightarrow \cos \theta=(\sqrt{2}-1) \sin \theta$

$\Rightarrow \frac{\cos \theta}{\sin \theta}=\sqrt{2}-1$

$\Rightarrow \cot \theta=\sqrt{2}-1$

The value of $\cot \theta$ is $\sqrt{2}-1$.

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Updated on: 10-Oct-2022

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