Can somone pleace prove this step by step answers ? (sinx)(cot x + cos x tan x)=cos x +sin^2x?

Answer 1

sin(x) [cot(x) + cos(x)tan(x)] = cos(x) + sin^2(x)

Choose the more complex side and try to get it to look like the other side. In this case, it's the left hand side.

LHS = sin(x) [cot(x) + cos(x)tan(x)]

Changing everything to sines and cosines, we have

LHS = sin(x) [ (cos(x)/sin(x)) + cos(x)[sin(x)/cos(x)] ]

Putting the terms in brackets under a common denominator,

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

Now that we have a product of terms, note the sin(x) outside of the brackets cancels with the sin(x) in the denominator within the brackets.

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

Now, we decompose this into two fractions; after all, we have an addition in the numerator. We decompose this in the same fashion that we can decompose (6 + 2)/2 as 6/2 + 2/2.

LHS = cos^2(x)/cos(x) + cos(x)sin^2(x)/cos(x)

Now, note the cancellations that occur. In the first fraction, the cos(x) in the denominator will cancel and the cos^2(x) in the numerator will have its power reduced by one. In the second fraction, the cos(x) on the top and bottom will cancel each other out. This leaves

LHS = cos(x) + sin^2(x) = RHS

Answer 2

by def. cot x=1/tg x = cos x/sin x and tg x is sinx/cosx

so the first member can be written as

sinx( cosx/sinx + cos x*sinx/cosx) and performing those operations

= cos x + sin^2x as you wanted

Answer 3

First thing to do is to remember that cotx = cosx/sinx and tanx = sinx/cosx

I will abreviate: sinx = s... cosx = c... tanx = s/c.... cot x = c/s

The problem is


s(c/s + cs/c) = c + s^2

s(c/s + s) = c + s^2

c + s^2 = c + s^2

Does all that make sense?

Answer 4

(sinx)(cot x + cos x tan x)=cos x +sin^2 x
LHS:
(sinx)(cot x + cos x tan x)
=sinx/tanx+sin^2 x
=cosx+sin^2 x
=RHS

there! (: