1) tan(sin^-1(5/6)-cos^-1(1/7)

2) tan(è + ); cos è = − 1/7,è in Quadrant III, sin = 1/2, in Quadrant II

Let A = sin^-1 (5/6) and B = cos^-1 (1/7)

then
tan(sin^-1(5/6)-cos^-1(1/7)
= tan(A-B)
= (tanA - tanB)/(1 + tanAtanB)

from A = sin^-1 (5/6)
sinA = 5/6, then cosA = √11/6 and tanA = 5/√11

from B = cos^-1 (1/7)
cosB = 1/7, then sinA = √48/7 and tanB = √48/1 = √48

then tan(A-B)
=(5/√11) - √48)/(1+(5/√11)(√48) )
= ( (5-√528)/√11) / (√11 + 5√48))/√11 )
= (5 - √528)/(√11 + 5√48)

= (5-4√33)/(11+20√3)

I cannot decipher your second question

To calculate the value of the given trigonometric expressions, we need to use the trigonometric identities and properties. Let's break down each expression step by step:

1) tan(sin^-1(5/6) - cos^-1(1/7))

First, we need to find the values of arcsin(5/6) and arccos(1/7).

- arcsin(5/6): This represents the angle whose sine value is equal to 5/6. Using the inverse sine function, we find the value of arcsin(5/6) as approximately 0.9273 radians or 53.13 degrees.

- arccos(1/7): This represents the angle whose cosine value is equal to 1/7. Using the inverse cosine function, we find the value of arccos(1/7) as approximately 1.4279 radians or 81.79 degrees.

Now, we can substitute these values into the given expression:

tan(sin^-1(5/6) - cos^-1(1/7))

= tan(0.9273 - 1.4279)

= tan(-0.5006)

Since the tangent function has periodicity of π, we can add or subtract multiples of π to the angle without changing the value of the tangent. Therefore, we can add π radians to -0.5006:

= tan(-0.5006 + π)

Now, we calculate the tangent:

= tan(2.641)

≈ 0.848

Therefore, the value of tan(sin^-1(5/6) - cos^-1(1/7)) is approximately 0.848.

2) tan(θ); cos(θ) = -1/7, θ in Quadrant III, sin(θ) = 1/2, in Quadrant II

We are given the values of cos(θ) and sin(θ), and we need to find the tan(θ).

Since θ is in Quadrant III and sin(θ) is positive, we know that θ is between π and 3π/2.

Since cos(θ) is negative, we can find the value of θ using arccos(1/7) and subtract from π:

θ = π - arccos(1/7)

Substituting the given value, we get:

θ ≈ π - 1.4279

Now, we can calculate tan(θ):

tan(θ) = tan(π - 1.4279)

Using the property tan(π - x) = -tan(x):

tan(θ) = -tan(1.4279)

Now, we calculate the tangent:

tan(θ) ≈ -0.942

Therefore, the value of tan(θ) is approximately -0.942.