tan[cos^-(-square root 3/2) + sin^-1(-4/5)]

Answer 1

The trigonometric expression is tan[cos^{- 1}(- √3/2) + sin^{- 1}(- 4/5)].

Let cos^{- 1}(- √3/2) = x  and  sin^{- 1}(- 4/5) = y.

Then, cos x = - √3/2  and  sin y = - 4/5

• If cos x = - √3/2, then

tan x =  √(1 - cos² x)/cos x.

= √(1 - (- √3/2)²)/(- √3/2)

= - [2√(1 - 3/4)]/√3

= - 2√(1/4)/√3

= - 2 * (1/2)/√3

= - 1/√3.

• If sin y = - 4/5, then

tan y =  sin y/√(1 - sin² y).

= (- 4/5)/√(1 - (- 4/5)²)

= - (4/5)/√(1 - 16/25)

= - (4/5)/√[(25 - 16)/25)]

= - (4/5)/√[9/25]

= - (4/5)/(3/5)

= - (4 * 5)/(5 * 3)

= - 4/3.

Apply formula : tan(x + y) = [tan x + tan y]/(1 - tan x tan y).

Substitute corresponding values.

tan(x + y) = [- 1/√3 - 4/3]/[1 - (- 1/√3)(- 4/3)]

= [(- 3 - 4√3)/3√3]/[(3√3 - 4)/3√3]

= [- 3 - 4√3]/[3√3 - 4]

Therefore, tan[cos^{- 1}(- √3/2) + sin^{- 1}(- 4/5)] = [- 3 - 4√3]/[3√3 - 4].