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

Explanation

Given

`f(x)=tanx`

Then`f'(x)=sec^2x`

Now,`f(x+Deltax)-f(x)=f'(x)Deltax`

`f(x+Deltax)-f(x)=(sec^2x)Deltax`

Taking `x=45^o=(pi/5)^o`

and `Delta=1^o 0.01745`

`f(46^o)-f(45)=(sec^2x)(0.01745)`

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