# Find the exact value of the expression: tan[tan^-1(-1)+sin^-1(-4/5)]....help please!

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To simplify this expression, use the "tangent of sum" identity:

`tan(x+y) = (tanx + tany)/(1-tanx*tany)`

In your expression, the first angle is `tan^(-1)(-1)` and the second angle is

`sin^(-1)(-4/5)` . Let's calculate the tangents of these angles separately and then plug them into the formula.

By the definition of the inverse tangent, tangent of the inverse tangent of -1 will be -1:

`tan(tan^(-1)(-1)) = -1` .

We cannot directly find the tangent of inverse sine, but we can find the sine of inverse sine:

`sin(sin^(-1)(-4/5)) = -4/5` .

Now we can express the tangent in terms of sine. Recall that

`tanx = (sinx)/(cosx)` and

according to the Pythagorean identity `cosx = sqrt(1-sin^2(x))` .

So the tangent of the angle with the sine equal to -4/5 will be

`tan(sin^(-1)(-4/5)) = (-4/5)/(sqrt(1-(-4/5)^2)) = (-4/5)/(3/5) = -4/3` .

So the tangents of the two angles in question are -1 and -4/3. Plugging these into the formula of the tangent of the sum, we get

`tan(tan^(-1)(-1) +sin^(-1)(-4/5)) = (-1 + (-4/3))/(1 - (-1)(-4/3)) = (-7/3)/(-1/3) = 7/1 = 7`

** So the exact value of the given expression is 7.**

Given to solve

`tan[tan^-1(-1)+sin^-1(-4/5)]`

we know ,

`tan^-1(-1) = (-pi)/4`

so ,

`tan[tan^-1(-1)+sin^-1(-4/5)] ` = `tan[(-pi)/4+(sin^-1(-4/5))]`

= `tan[sin^-1(-4/5) - (pi)/4]` -----(1)

so the above expression is of the form`tan(a-pi/4)` , we know

`tan(a-pi/4) = (tan a -1 )/(1+tan a)`

so **applying the above formula we** , get

`tan[sin^-1(-4/5) -(pi)/4]`

= `(tan (sin^-1(-4/5))-1 )/(1+tan (sin^-1(-4/5)))` ----(2)

**let** `sin^-1(-4/5) =a`

=>`sin a = -4/5`

=> Using all silver tea cups property ( see the attachment) we get ,

`tan a = -4/3`

=> `a = tan^-1(-4/3) ` so we can say that

**`sin^-1(-4/5) = a = tan^-1(-4/3)` -----(3)**

Now substituting the value of (3) in (2) , we get

`(tan(sin^-1(-4/5))-1 )/(1+tan(sin^-1(-4/5)))`

= `(tan(tan^-1(-4/3))-1 )/(1+tan(tan^-1(-4/3)))`

= `((-4/3)-1 )/(1+(-4/3))`

= `((-7)/(3))/((-1)/3)`

= ` 7`