Answer
456k+ views
Hint: - Use trigonometric identity ${\text{cos}}\left( {90^\circ - \theta } \right) =
\sin \theta $
As given in the question let’s first solve the given expression: -
$ \Rightarrow \sin \theta + \cos \theta = \sqrt 2 \cos \left( {90^\circ - \theta } \right)$
Therefore, above expression will become
$ \Rightarrow \sin \theta + \cos \theta = \sqrt 2 \sin \theta {\text{ }}$ as we know
${\text{cos}}\left( {90^\circ - \theta } \right) = \sin \theta $
$
\Rightarrow {\text{so,cos}}\theta {\text{ = }}\left( {\sqrt 2 - 1} \right)\sin \theta \\
\Rightarrow \cot \theta = \sqrt 2 - {\text{1}}{\text{.}} \\
$
Note: - Whenever this kind of question appears always first simplify the equation as much as
possible. Remember in this type of question basic knowledge of trigonometric identities is
must. Remember $\cos \theta $ always remain positive in the fourth quadrant.
\sin \theta $
As given in the question let’s first solve the given expression: -
$ \Rightarrow \sin \theta + \cos \theta = \sqrt 2 \cos \left( {90^\circ - \theta } \right)$
Therefore, above expression will become
$ \Rightarrow \sin \theta + \cos \theta = \sqrt 2 \sin \theta {\text{ }}$ as we know
${\text{cos}}\left( {90^\circ - \theta } \right) = \sin \theta $
$
\Rightarrow {\text{so,cos}}\theta {\text{ = }}\left( {\sqrt 2 - 1} \right)\sin \theta \\
\Rightarrow \cot \theta = \sqrt 2 - {\text{1}}{\text{.}} \\
$
Note: - Whenever this kind of question appears always first simplify the equation as much as
possible. Remember in this type of question basic knowledge of trigonometric identities is
must. Remember $\cos \theta $ always remain positive in the fourth quadrant.
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