We have to prove tan*sin *(cos+1) + cos x = sin^2 x + sec x

tan*sin *(cos+1) + cos x

use tan x = sin x/ cos x

=> (sin x/cos x)*sin x * (cos x + 1) + cos x

=> (sin x)^2/ cos x * (cos x + 1) + cos x

=> [(sin x)^2* (cos x) + (sin x)^2 + (cos x)^2]/ cos x

=> [(sin x)^2* (cos x) + 1]/ cos x ...(1)

(sin x)^2 + sec x

use sec x = 1/cos x

=> (sin x)^2 + sec x

=> (sin x)^2 + 1/cos x

=> [(sin x)^2*cos x + 1]/cos x ...(2)

It can be seen that (1) and (2) have the same value.

**This proves that tan*sin *(cos+1) + cos x = sin^2 x + sec x**

tanx*sinx* (cosx+1) + cosx = sin62 x+ sec x

Let us start from the left side and prove the right side.

We know that tanx = sinx/cosx

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

==> We will rewrite using the common denominator.

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

==> (sin^2 x * cosx + sin62 x + cos^2 x) / cosx

But sin^2 x + cos^2 x = 1

==> (sin^2 x *cosx + 1 ) / cosx

==> sin^2 x cosx/ cosx + 1/cosx

==> sin^2 x + sec x .............q.e.d

**Then we have prove that : tanx*sinx(cosx+1) + cosx = sin^2 x + sec x**