Question

x=(tanA+cotA)^2 sinA-tan^2A

Answer 1

\[x=(tanA+cotA)^2 *(sinA-\tan^2A)\]This?

Answer 2

is @henpen right?

Answer 3

yes

Answer 4

\[(tanA+cotA)=(\frac{sinA}{cosA}+\frac{cosA}{sinA})=(\frac{sinAsinA}{cosAsinA}+\frac{cosAcosA}{sinAcosA})=\frac{1}{cosAsinA}\]

Answer 5

\[\because \cos^2(x)+\sin^2(x)=1\]

Answer 6

\[sinA-\tan^2A=\frac{sinAcos^2A}{\cos^2A}-\frac{\sin^2A}{\cos^2A}=\frac{sinAcos^2A-\sin^2A}{\cos^2A}\]

Answer 7

give me a minute to analyze what you had illustrated

Answer 8

Multiplying,\[(\frac{1}{cosAsinA})^2*(\frac{sinAcos^2A-\sin^2A}{\cos^2A})=\frac{sinAcos^2A-\sin^2A}{\cos^4Asin^2A}\]

Answer 9

Which you can simplify if you want.

Answer 10

you must get this csc(x)-sec^2(x)

Answer 11

owh.. i'm sorry .. you should find the value of x?
the possible answers are:
a. 4 b.3 c.2 d.1

Answer 12

\[cscAsec^2A-\sec^4A\]I must be mistaken.

Answer 13

how will i arrive at the possible answers?