Is sin(arcsin(x)+arccos(x))=1 ?

The answer to whether sin(arcsin(x)+arccos(x))=1 is True

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prove sin(arcsin(x)+arccos(x))=1

prove

sin (arcsin (x) + arccos (x)) = 1

True

Solution steps

sin (arcsin (x) + arccos (x)) = 1

Manipulating left side

sin (arcsin (x) + arccos (x))

Rewrite using trig identities

sin (arcsin (x) + arccos (x))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (arcsin (x)) cos (arccos (x)) + cos (arcsin (x)) sin (arccos (x))

sin (arcsin (x)) cos (arccos (x)) + cos (arcsin (x)) sin (arccos (x)) = 1

sin (arcsin (x)) cos (arccos (x)) + cos (arcsin (x)) sin (arccos (x))

Use the following identity:

cos (arccos (x)) = x

= sin (arcsin (x)) x + cos (arcsin (x)) sin (arccos (x))

Use the following identity:

cos (arcsin (x)) = 1 - x^{2}

= sin (arcsin (x)) x + 1 - x^{2} sin (arccos (x))

Use the following identity:

sin (arcsin (x)) = x

= xx + 1 - x^{2} sin (arccos (x))

Use the following identity:

sin (arccos (x)) = 1 - x^{2}

= xx + 1 - x^{2} 1 - x^{2}

Simplify

xx + 1 - x^{2} 1 - x^{2}

xx = x^{2}

xx

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

xx = x^{1 + 1}

= x^{1 + 1}

Add the numbers:

1 + 1 = 2

= x^{2}

1 - x^{2} 1 - x^{2} = 1 - x^{2}

1 - x^{2} 1 - x^{2}

Apply radical rule:

a a = a

- x^{2} + 1 - x^{2} + 1 = 1 - x^{2}

= 1 - x^{2}

= x^{2} + 1 - x^{2}

Group like terms

= x^{2} - x^{2} + 1

Add similar elements:

x^{2} - x^{2} = 0

= 1

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 - 2 sec ( x ) - 3 sec ^{2} ( x )}{- tan ^{2} ( x )} = \frac{1 - 3 sec ( x )}{1 - sec ( x )}

\frac{1}{2 csc ( 2 x )} = cos^{2} (x) tan (x)

(sec (5 x) - 1) (sec (5 x) - 1) = tan^{2} (5 x)

1 - cos (x) sin (x) cot (x) = sin^{2} (x)

cot (x) + tan (x) = sec (x) \cdot csc (x)

Is sin(arcsin(x)+arccos(x))=1 ?
The answer to whether sin(arcsin(x)+arccos(x))=1 is True