If $\sin \left( {90 - heta } ight) + \cos heta = \sqrt 2 \cos \left( {90 - heta } ight),$for $0 $ { ext{A}}{ ext{. }}\frac{{\sqrt 3 }}{2} \ { ext{B}}{ ext{. }}\frac{2}{{\sqrt 3 }} \ { ext{C}}{ ext{. }}\sqrt {\frac{3}{2}} \ { ext{D}}{ ext{. }}\sqrt {\frac{2}{3}} \ $

Question

If $\sin \left( {90 - 	heta } ight) + \cos 	heta  = \sqrt 2 \cos \left( {90 - 	heta } ight),$for $0 $  {	ext{A}}{	ext{. }}\frac{{\sqrt 3 }}{2}   \  {	ext{B}}{	ext{. }}\frac{2}{{\sqrt 3 }}   \  {	ext{C}}{	ext{. }}\sqrt {\frac{3}{2}}    \  {	ext{D}}{	ext{. }}\sqrt {\frac{2}{3}}    \ $

Vedantu Content Team · Accepted Answer

Hint: - For solving this type of question you have just knowledge of trigonometric transformation and just simple mathematics to proceed further. You have to transform as you can shorten the complex equation in easy form.

Complete step-by-step solution -

As given in question
$\sin \left( {90 - \theta } \right) + \cos \theta = \sqrt 2 \cos \left( {90 - \theta } \right)$
We know $\left( {\because \sin \left( {90 - \theta } \right) = \cos \theta } \right)\left( {\because \cos \left( {90 - \theta } \right) = \sin \theta } \right)$
$ \Rightarrow \cos \theta + \cos \theta = \sqrt 2 \sin \theta $
$ \Rightarrow 2\cos \theta = \sqrt 2 \sin \theta $$ \Rightarrow \frac{{\cos \theta }}{{\sin \theta }} = \frac{{\sqrt 2 }}{2} = \frac{1}{{\sqrt 2 }}$
We know $\left( {\frac{{\cos \theta }}{{\sin \theta }} = \cot \theta } \right)$
$ \Rightarrow \cot \theta = \frac{1}{{\sqrt 2 }}$ $\left( {\because \cos e{c^2}\theta - {{\cot }^2}\theta = 1} \right)\left( {\therefore \cot \theta = \sqrt {\cos e{c^2}\theta - 1} } \right)$
$ \Rightarrow \sqrt {\cos e{c^2}\theta - 1} = \frac{1}{{\sqrt 2 }}$
Squaring on both side, we get
$ \Rightarrow \cos e{c^2}\theta - 1 = \frac{1}{2}$
$\therefore \cos e{c^2}\theta = \frac{3}{2} \Rightarrow \cos ec\theta = \sqrt {\frac{3}{2}} $
Hence the option ${\text{C}}$is the correct option.

Note: -Whenever you get these types of questions the key concept of solving is you have to proceed from the question and just use trigonometric results to get an answer. You have to shorten the complex equation using standard results.

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