Question

Question #d325d

Answer 1

To be able to add these together we will use cross multiply to get the fractions to have similar denominators.

`a/b+c/d=(a*d)/(b*d)+(c*b)/(b*d)=(ad+cb)/(bd)`

Applying this, we get:

`((1+sinx)/cosx)+(cosx/(1+sinx))`

`=((1+sinx) * (1+sinx))/(cosx * (1+sinx))+(cosx * cosx)/(cosx * (1+sinx))`

`=((1+sinx)(1+sinx)+(cosx)(cosx))/(cosx(1+sinx))`

`=(1+2sinx+sin^2x+cos^2x)/(cosx(1+sinx))`

We can use the identity that `cos^2x+sin^2x-=1` to get:

`=(1+2sinx+1)/(cosx(1+sinx))`

`=(2+2sinx)/(cosx(1+sinx))`

`=(2(1+sinx))/(cosx(1+sinx))`

`=(2cancel((1+sinx)))/(cosxcancel((1+sinx)))`

`=2/cosx`

`=2secx`

Answer 2

`"see explanation"`

Explanation:

`"consider the left side"`

`"we require the fractions to have a "color(blue)"common denominator"`

`"multiply numerator/denominator of"`

`(1+sinx)/cosx" by "(1+sinx)`

`rArr(1+sinx)^2/(cosx(1+sinx))`

`"multiply numerator/denominator of"`

`cosx/(1+sinx)" by "cosx`

`rArrcos^2x/(cosx(1+sinx))`

`"we now have the sum"`

`(1+sinx)^2/(cosx(1+sinx))+cos^2x/(cosx(1+sinx))`

`"expand and sum the numerators"`

`=(1+2sinx+sin^2x+cos^2x)/(cosx(1+sinx))`

`•color(white)(x)sin^2x+cos^2x=1`

`=(2cancel((1+sinx)))/(cosxcancel((1+sinx)))`

`=2/cosx=2secx=" right side "rArr" verified"`