Question

The two legs of right triangle are $\sin \theta +\sin \left(\frac{3\pi }{2}-\theta \right)$ and $\cos \theta -\cos \left(\frac{3\pi }{2}-\theta \right)$. The length of its hypotenuse is

• A. $1$
• B. $\sqrt{2}$
• C. $2$
• D. some function of $\theta$

Answer 1

Answer: (B)

$\sqrt{2}$

Let $sin\theta +sin(\frac{3\pi }{2}-\theta )=sin\theta -cos\theta =a$

and

Let $cos\theta -cos(\frac{3\pi }{2}-\theta )$$=cos\theta +sin\theta$$=b$

Hence

$c^2=a^2+b^2$ ... by Pythagoras theorem.
$c^2=(sin\theta -cos\theta {)}^2+(sin\theta +cos\theta {)}^2$

$c^2=2(si{n}^2\theta +co{s}^2\theta )$

Thus

$c^2=2$

$\Rightarrow c=\sqrt{2}$