Question

For x such that 0 < x < ~, the expression V1 - cos x + V1 - sin x is equivalent to: sin x COS X F. 0 G. 1 H. 2 J. -tan x K. Sin 2x

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is H

Step-by-step explanation:

From the question we are told that

   The equation is  $\frac{\sqrt{1 - cos^2 (x)} }{sin(x)} + \frac{\sqrt{1 - sin^2 (x)} }{cos(x)}$    

    The domain for x  is  $0 < x < \frac{\pi}{2}$

Gnerally the equation above is not continuous, when

       $sin (x) = 0$

=>    $x = 0$

And  when  $cos(x) = 0$

           =>     $x = \frac{\pi}{2}$

Generally  from trigonometry identity

        $sin^2x + cos^2 x = 1$

So    

       $sin^2 x = 1 - cos^2 (x)$

So

      $cos^2 x = 1 - sin^2 (x)$    

=>     $\frac{\sqrt{sin^2 (x)} }{sin(x)} + \frac{\sqrt{ cos^2 (x )} }{cos(x)}$    

=>     $1 + 1$      

=>     $2$