Question

How to prove 5secx+4 =9secsquarex- 4tansquarex?

`5sec^2x+4 =9sec^2x- 4tan^2x?`

Answer 1

Please find a Proof in Explanation.

Explanation:

I hope, the Question is to prove :

`5sec^2x+4 =9sec^2x- 4tan^2x.`

If we use the Identity : `tan^2x=sec^2x-1`, the Problem

can be easily proved as follows :

`9sec^2x-4tan^2x`,

`=9sec^2x-4(sec^2x-1)`,

`=9sec^2x-4sec^2x+4`,

`=5sec^2x+4`, as desired!

Otherwise, `9sec^2x-4tan^2x`,

`=9sec^2x-4{(sin^2x)/cos^2x}`,

`=9sec^2x-4{(1-cos^2x)/cos^2x}`,

`=9sec^2x-4{1/cos^2x-cos^2x/cos^2x}`,

`=9sec^2x-4{sec^2x-1}`, and the rest is as above.

Answer 2

The question is in error, and in fact

`5sec^2x+4 -= 9sec^2x-4tan^2x`

Explanation:

The question is in error, and in fact

`5sec^2x+4 -= 9sec^2x-4tan^2x`

As can be shown, if we consider the LHS:

`LHS -= 5sec^2x+4`

`\ \ \ \ \ \ \ \ = 5(1+tan^2x)+4`

`\ \ \ \ \ \ \ \ = 5+5tan^2x+4`

`\ \ \ \ \ \ \ \ = 9+5tan^2x`

`\ \ \ \ \ \ \ \ = 9+5tan^2x + 4tan^2x - 4tan^2x`

`\ \ \ \ \ \ \ \ = 9+9tan^2x - 4tan^2x`

`\ \ \ \ \ \ \ \ = 9(1+tan^2x) - 4tan^2x`

`\ \ \ \ \ \ \ \ = 9sec^2x - 4tan^2x`

`\ \ \ \ \ \ \ \ -= RHS \ \ \ \ \` QED

Answer 3

Answer for the question :
`5color(red)(sec^2x)+4=9sec^2x-4tan^2x`
Please see below

Explanation:

We know that,

#color(blue)(sec^2theta-tan^2theta=1=>sec^2theta- 1=tan^2theta#

We have to prove,

`5sec^2x+4=9sec^2x-4tan^2x`

We take,

`RHS=9sec^2x-4tan^2x`

`color(white)(RHS)=9sec^2x-4(sec^2x-1)`

`color(white)(RHS)=9sec^2x-4sec^2x+4`

`color(white)(RHS)=5sec^2x+4`

`RHS=LHS`