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\theta } \right)$, then find the value of ${\text{cot}}\theta $.

$

{\text{A}}{\text{. }}\frac{1}{2} \\

{\text{B}}{\text{. 0}} \\

{\text{C}}{\text{. }}\sqrt 2 - 1 \\

{\text{D}}{\text{. }}\sqrt 2 \\

$

Answer

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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.