simplify trig expression?????

Question

csc^2(θ) / (1 + tan^2(θ) )

Answer 1

1 + tan²x = sec²x, so you have

csc²x/sec²x
= 1/sin²x / 1/cos²x
= cos²x/sin²x
= cot²x

Answer 2

The Pythagorean Identity with tangent in it is: 1 + tan^2 x = sec^2 x.

csc^2(x) / (1 + tan^2(x) )
csc^2(x) / sec^2(x)

So, that is the inverse of tan^2 x, or
cot^2(x)

Answer 3

sec^2(x) is 1+tan^2(x)
This comes from the trig identity sin^2(x)+cos^2(x)=1 and dividing through by cos^2(x)

Hence expression is now cosec^2(x)/sec^2(x)
cosec(x)=1/sin(x) and sec(x)=1/cos(x)
Hence the expression becomes cos^2(x)/sin^2(x) which is
1/tan^2(x)

Hence your answer is tan^-2(x) or 1/tan^2(x), cot^2(x)

Answer 4

use the identity sec^2(a)=1+tan^2(a)

Answer 5

Do with the aid of fact the previous answerers propose and get: (sinx +sinx/cosx +a million +a million/cosx) / (a million/cosx +a million) Multiply numerator and denominator by technique of using cosx: sinxcosx +sinx +cosx +a million) / (a million +cosx) = sinx(cosx +a million) +a million(cosx +a million) / (a million+cosx) = (cosx+a million)(sinx +a million) / (a million+cosx) = sinx +a million

Answer 6

csc^2(θ) / (1 + tan^2(θ) )

**
1 + tan^2θ = sec^2θ because:
LHS = (cos^2θ + sin^2θ)/cos^2θ
= 1/cos^2θ
= sec^2θ
**
Sub this into formula gives us:
csc^2θ/sec^2θ

= cos^2θ/sin^2θ

= cot^2θ

Hope this helps!

Answer 7

let x = theta:-
(1 / sin ² x) / (1 + sin ² x / cos ² x)
(cos ² x / sin ² x) / (cos ² x + sin ² x)
cos ² x / sin ² x
cot ² x