Using differentials find the approximate value of `tan46^0,` if it is being given that `1^0=0. 01745` radians.

Explanation

Let `f(x)=tanx`.
Then, `f'(x)=sec^2x`

Now,
`f(x+δx)−f(x)=f'(x).δx`

`⇒f(x+δx)−f(x)=sec^2x.δx...(i)`

Putting `x=45^o,δx=1^o=0.01745` in `(i)`, we get

`f(46^o)−f(45^o)=(sec^(2)45^o)×0.01745`

`⇒tan46^o−tan45^o=(sec^(2)45^o)×0.01745`

`⇒tan46^o=tan45^(o)+(sec^(2)45^o)×0.01745`

`=(1+2×0.01745)`
`=1.03490`

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