I need to solve `tan(cos^-1(-4/5) + csc^-1((-sqrt 41)/5))` without using a calculator and show the working out.

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durbanville | High School Teacher | (Level 2) Educator Emeritus

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You need to use compound angles to solve. We know that

`tan(a+b)= (sin (a+b))/(cos (a+b))`

`therefore = (sina cosb + cos a sinb)/(cosa cosb - sina sinb)`

We know, from the question that `a = cos^(-1)((-4)/5)` and `b=csc^(-1)((-sqrt41)/5)`   `therefore = sin(cos^(-1)((-4)/5)). cos (csc^(-1)(-sqrt41/5))+ cos(cos^(-1)(-4/5)). sin(csc^(-1)((-sqrt41)/5)`

represents the numerator and for the denominator:

`cos(cos^-1 (-4/5)). cos(csc^-1 ((-sqrt41)/5)) - sin(cos^-1 ((-4)/5)). sin(csc^-1 ((-sqrt41)/5))`

Simplify numerator and denominator separately to avoid confusion. In the numerator we need to change sin to cos  and cos to sin first so we can simplify the first part of term 1. The 2nd part (after the + in the numerator) is already cos with cos and csc with sin. As `sin^2 = 1-cos^2` .`therefore sin=sqrt(1-cos^2)`

and the same for cos where `cos^2 = 1-sin^2` . Also note that  `csc = 1/sin`

`therefore` `sqrt(1-cos^2)(cos^(-1)((-4)/5)). sqrt(1-sin^2)((1/sin)^(-1)((-sqrt41)/5))` `+(-4)/5. 5/(-sqrt41)`

`therefore 3/5 . 4/sqrt41 + (-20)/(-5sqrt41)`

`= 32/(5sqrt41)` (numerator)

and therefore when solving the denominator we already know that `cos a = (-4)/5` and `cos b=4/sqrt41` and we know that ` sin a=3/5` and `sin b=5/-sqrt41`

from having calculated the numerator.

`therefore = -4/5. 4/sqrt41 - 3/5. 5/(-sqrt41)`

`therefore = -1/(5sqrt41)`

Now put numerator over denominator:

`32/(5sqrt41) divide -1/(5sqrt41)`

= -32

Ans: -32

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