Question

How do you evaluate [sin 70 degrees/cos(-430 degrees)] + tan(-70 degrees)?

Answer 1

Explanation is given below

Explanation:

Before we try and evaluate let us go over these identities.

`cos(360n+theta) = cos(theta)`
`cos(-theta) = cos(theta)`
`sin(theta)/cos(theta) = tan(theta)`
`sin(-theta) = -sin(theta)`
`tan(-theta) = -tan(theta)`

Now let us come to our problem

`sin(70^o)/cos(-430^o) + tan(-70^o)`

`=sin(70^o)/cos(430^o)-tan(70^o)` `quad color(red) "since"quad tan(-theta)=tan(theta) and cos(-theta)=cos(theta)`

`=sin(70^o)/cos(360^o +70^o)-tan(70^o)` `quad color(red) "since"quad 430=360+70`

`=sin(70^o)/cos(70^o) - tan(70^o)` `quad color(red) "since"quad cos(360+theta)=cos(theta)`

`=tan(70^o)-tan(70^o)` `quad color(red) "since"quad sin(theta)/cos(theta) = tan(theta)`

`=0` Answer