Question

@ganeshie8

Answer 1

\(\sf cotx~sec^4x=cotx+2tanx+tan^3x\)

Answer 2

Decide which side you want to start with

Answer 3

If it doesn't matter which side then I guess \(\sf cotx+2tanx+tan^3x\)

Answer 4

thats a good start!
notice that there is a factor \(\cot x\) on left hand side so does it make sense to factor out \(\cot x\) from the right hand side also ?

Answer 5

Right hand side :
\[\sf cotx+2tanx+tan^3x\]

factoring out \(\cot x\) you get :
\[\sf cotx \left (1 +\frac{2tanx}{\cot x}+\frac{tan^3x}{\cot x}\right)\]

Answer 6

ooohhhh that makes sense

Answer 7

next use the fact that \(\frac{1}{\cot x} = \tan x\)

Answer 8

Right hand side :
\[\sf cotx+2tanx+tan^3x\]

factoring out \(\cot x\) you get :
\[\sf cotx \left (1 +\frac{2tanx}{\cot x}+\frac{tan^3x}{\cot x}\right)\]

rewriting 1/cot as tan and simplifying gives :

\[\sf cotx \left (1 +2tan^2x+tan^4x\right)\]

Answer 9

we're almost done!
the next step is to complete the square by using a well known identity
\[a^2 + 2ab + b^2 = (a+b)^2\]

Answer 10

see if you can cleverly apply that identity to the stuff inside parenthesis

Answer 11

I think it would have to become \(\sf (1+2tanx+tan^2x)^2\) first.

Answer 12

Right hand side :
\[\sf cotx+2tanx+tan^3x\]

factoring out \(\cot x\) you get :
\[\sf cotx \left (1 +\frac{2tanx}{\cot x}+\frac{tan^3x}{\cot x}\right)\]

rewriting 1/cot as tan and simplifying gives :

\[\sf cotx \left (1 +2tan^2x+tan^4x\right)\]


\[\sf cotx \left (1+\tan^2 x\right)^2\]


just expand that and convince yourself that you get the previous expression back :)

Answer 13

\(\sf cot(1+2tan^2x+tan^4x)\)

Answer 14

thats same as previous expression. so we are good

Answer 15

see if u can figure out the final step

Answer 16

Distribute cot
\(\sf cot+2tan^2x~cot+tan^4x~cot\)

Answer 17

No, then we will be going in reverse direction :O

Answer 18

:(

Answer 19

look at the left hand side
you want sec^4x term yes ?

Answer 20

I will need to use the same identity as last time?
\(\sf 1+tan^2x=sec^2x\)

Answer 21

BINGO!

Answer 22

\(\sf cot(2sec^2x+tan^4x)\)?

Answer 23

nope, so far we have : \[\sf cotx \left (1+\tan^2 x\right)^2\]

use the identity for stuff inside parenthesis
\[\sf cotx \left (\sec^2(x)\right)^2\]
conclude
\[\sf cotx~ \sec^4(x)\]

Answer 24

ooohhh ok, that makes more sense than what I was thinking

Answer 25

we need to keep in mind what our goal is
our goal is the left hand side : cotx sec^4x

Answer 26

Btw you might be knowing this notation thing :
\[\tan^2(x)\]

is same as

\[\left[\tan (x) \right]^2\]

Answer 27

Yep, I realized that after my 'lessons' used them both a few times.

Answer 28

yeah i remember thats one of the things that use to confuse the hell outof me in trig but i see you're doing fine :)