Question

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

• A. $\sqrt{2}sec({90}^{\circ }-\theta )$
• B. $\sqrt{2}sin({90}^{\circ }-\theta )$
• C. $\sqrt{2}sin\theta$
• D. $\sqrt{2}cos\theta$

Answer 1

The correct option is C

$\sqrt{2}sin\theta$

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

$cos\theta +sin\theta =\sqrt{2}cos\theta$ ....(i)

We know

$(cos\theta +sin\theta {)}^2+(cos\theta -sin\theta {)}^2=2(co{s}^2\theta +si{n}^2\theta )$

$(\sqrt{2}cos\theta {)}^2+(cos\theta -sin\theta {)}^2=2....(co{s}^2\theta +si{n}^2\theta =1)$

$2co{s}^2\theta +(cos\theta -sin\theta {)}^2=2$

$(cos\theta -sin\theta {)}^2=2-2co{s}^2\theta$

$(cos\theta -sin\theta {)}^2=2(1-co{s}^2\theta )$

$(cos\theta -sin\theta {)}^2=2si{n}^2\theta$

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