Prove that arcsinx + arccosx = pi/2 .

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Take a right triangle with the hypotenuse equal to 1.

Now let one of the other sides be equal to x. If the third side is s, s^2 + x^2 = 1

=> s^2 = 1 - x^2

=> s = sqrt (1- x^2)

Now for the triangle, if the...

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Take a right triangle with the hypotenuse equal to 1.

Now let one of the other sides be equal to x. If the third side is s, s^2 + x^2 = 1

=> s^2 = 1 - x^2

=> s = sqrt (1- x^2)

Now for the triangle, if the angles are a and b.

sin(a) = x

cos(a) = sqrt(1 - x^2)

sin(b) = sqrt(1 - x^2)

cos(b) = x

Now, sin(a + b)

=> sin(a)cos(b) + cos(b)sin(a)

=> x^2 + (1-x^2)

=> 1

Also, sin (pi/2) = 1.

So we have a + b = pi/2

But a = arc sin (x) and b = arc cos (x).   

Therefore arc sin (x) + arc cos (x) = pi/2

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