What is the value of tan(arcsin(4/5)+arccos(1)) ?

The value of tan(arcsin(4/5)+arccos(1)) is 4/3

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tan(arcsin(4/5)+arccos(1))

tan (arcsin (\frac{4}{5}) + arccos (1))

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

tan (arcsin (\frac{4}{5}) + arccos (1))

Rewrite using trig identities:

\frac{tan ( arcsin ( \frac{4}{5} ) ) + tan ( arccos ( 1 ) )}{1 - tan ( arcsin ( \frac{4}{5} ) ) tan ( arccos ( 1 ) )}

tan (arcsin (\frac{4}{5}) + arccos (1))

Use the Angle Sum identity:

tan (s + t) = \frac{tan ( s ) + tan ( t )}{1 - tan ( s ) tan ( t )}

= \frac{tan ( arcsin ( \frac{4}{5} ) ) + tan ( arccos ( 1 ) )}{1 - tan ( arcsin ( \frac{4}{5} ) ) tan ( arccos ( 1 ) )}

= \frac{tan ( arcsin ( \frac{4}{5} ) ) + tan ( arccos ( 1 ) )}{1 - tan ( arcsin ( \frac{4}{5} ) ) tan ( arccos ( 1 ) )}

Rewrite using trig identities:

tan (arcsin (\frac{4}{5})) = \frac{4}{3}

tan (arcsin (\frac{4}{5}))

Rewrite using trig identities:

tan (arcsin (\frac{4}{5})) = \frac{( \frac{4}{5} ) 1 - ( \frac{4}{5} ) ^{2}}{1 - ( \frac{4}{5} ) ^{2}}

Use the following identity:

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

= \frac{( \frac{4}{5} ) 1 - ( \frac{4}{5} ) ^{2}}{1 - ( \frac{4}{5} ) ^{2}}

= \frac{\frac{4}{5} 1 - ( \frac{4}{5} ) ^{2}}{1 - ( \frac{4}{5} ) ^{2}}

Simplify

= \frac{4}{3}

Rewrite using trig identities:

tan (arccos (1)) = 0

tan (arccos (1))

Rewrite using trig identities:

tan (arccos (1)) = \frac{1 - 1 ^{2}}{1}

Use the following identity:

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

= \frac{1 - 1 ^{2}}{1}

= \frac{1 - 1 ^{2}}{1}

Simplify

= 0

= \frac{\frac{4}{3} + 0}{1 - \frac{4}{3} \cdot 0}

Simplify

= \frac{4}{3}

\frac{2 tan ( \frac{7 π}{8} )}{1 - tan ^{2} ( \frac{7 π}{8} )}

sin (11 3^{\circ})

arctan (0.52)

arctan (0.89)

sin (2 0^{\circ}) cos (8 0^{\circ}) - cos (2 0^{\circ}) sin (8 0^{\circ})

What is the value of tan(arcsin(4/5)+arccos(1)) ?
The value of tan(arcsin(4/5)+arccos(1)) is 4/3