Can someone please help me with the following trig identity: -(csc^2ø)(cot^2ø) + cot^2ø = -cot^4ø !?

Answer 1

I'm going to use t instead of that symbol.

-[csc^2(t)] [cot^2(t)] + cot^2(t) = -cot^4(t)

As per the usual steps in solving identities, you need to choose the more complex side. In this case, it's the left hand side.

LHS = -[csc^2(t)] [cot^2(t)] + cot^2(t)

Convert them all to sines and cosines.

LHS = -[ 1/sin^2(t) ] [cos^2(t) / sin^2(t)] + [cos^2(t) / sin^2(t)]

Multiplying the fractions out, to get

LHS = [-cos^2(t) / sin^4(t)] + [cos^2(t) / sin^2(t)]

Putting them under a common denominator of sin^4(t), we get

LHS = [-cos^2(t) / sin^4(t)] + [sin^2(t)][cos^2(t)] / [sin^4(t)]

Merging them into one fraction,

LHS = (-cos^2(t) + [sin^2(t)][cos^2(t)]) / sin^4(t)

Factoring the numerator by pulling out -cos^2(t),

LHS = -cos^2(t) [1 - sin^2(t)] / [sin^4(t)]

Note that 1 - sin^2(t) is a familiar identity; I'll leave that up to you to look up, if you don't know it already.

LHS = -cos^2(t) [cos^2(t)] / sin^4(t)

Simplifying,

LHS = -cos^4(t)/sin^4(t)

And now, by definition, since

LHS = - [cos^4(t) / sin^4(t)]

We have

LHS = -cot^4(t) = RHS

Answer 2

I put t for theta

LHS = - csc^2 t cot^2 t + cos ^2 t
= cot^2 t( 1- csc^2 t)

but csc^2 t = 1+ cot ^2 t
so LHS = cot^2 t(1-(1+cot^2 t) = cot^2 t *(-cot^2 t) = - cot^4 t = RHS

Answer 3

Remember the following and you will get most of trigo.

sin x = 1/cosec x
cos x = 1/sex x
tan x = 1/cot x

sin^2 x + cos^2 x = 1
sec^2 x - tan^2 x = 1
cosec^2 x - cot^2 x = 1

Answer 4

.

Firstly let me use "x" instead of "ø" for simplicity.

I hope you have learnt the 3 basic Trigonometric identities:

sin^2(x) + cos^2(x) = 1
1 + tan^2(x) = sec^2(x)
1 + cot^2(x) = cosec^2(x)

If you have not, I hope you are aware of Pythagoras' Theorem, which gives:

(hypotenuse)^2 = (side1 of right angle)^2 + (side2 of right angle)^2
Let the above equation be called (A)

sine of an angle = side opposite to the angle / hypotenuse
cosine of an angle = side adjacent to the angle / hypotenuse

Now, dividing (A) by (hypotenuse)^2, we will get:
1 = (side1/hypotenuse)^2 + (side2/hypotenuse)^2

This is nothing but
sin^2(x) + cos^2(x) = 1.

We have thus arrived at (i). Equations (ii) and (iii) can be derived from (i) by simply dividing by cos^2(x) and sin^2(x) respectively.


Now coming to your problem...

To Prove: cosec^2(x) * cot^2(x) + cot^2(x) = cot^4(x)

Let us consider the Left Hand Side of the equation and try to arrive at the Right Hand Side of the equation...

cosec^2(x) * cot^2(x) + cot^2(x)

cot^2(x) is a common factor of the 2 terms...
cot^2(x) * [ cosec^2(x) + 1 ]

Using Identity (iii)...
cot^2(x) * [cot^2(x)]

Which reduces to the RHS of the equation,
cot^4(x)

Hence Proved.

.

Answer 5

We have to prove

-(csc^2ø)(cot^2ø) + cot^2ø = -cot^4ø

Let (cot^2ø) = y.

We have to prove

-(csc^2ø)y + y= - y^2.

Or

y { 1- csc^2ø } = - y^2.

Or

{ 1 - csc^2ø } = - y.

Since 1 - csc^2ø = - cot^2ø = -y.

it is proved.

Answer 6

- csc² ø cot² ø + cot² ø = -cot^4 ø

We can factor out cot² ø on the left side
cot² ø(-csc² ø + 1) = -cot^4 ø

We rearrange the terms inside the parentheses
cot² ø(1 - csc² ø) = -cot^4 ø

Since csc² ø = 1 + cot² ø,
cot² ø[1 - (1 + cot² ø)] = -cot^4 ø

We combine like terms
cot² ø(-cot² ø) = -cot^4 ø

Therefore,
-cot^4 ø = -cot^4 ø

QED
^_^

Answer 7

Let's work with the left hand side of the identity.

-(csc² ø)(cot² ø) + cot² ø = -cot^4 ø
= (cot² ø){-1/sin² ø + 1)
= (cot² ø){(-1 + sin² ø)/sin² ø)
= (cot² ø){-cos² ø/sin² ø)
= (cot² ø)(-cot² ø)
= -cot^4 ø

qed

Answer 8

for solving that u have to use these formulas:
1)sin^2(x)+cos^2(x)=1
2)cosec^2(x)=1/(sin^2(x))
3)cotg(x)=cos(x)/sin(x)
first step:we use number (2) so we have:

=(-1/sin^2(x))*cotg^2(x)+cotg^2(x) now use number (1)
=((-sin^2(x)-cos^2(x))/sin^2(x))*cotg^2(x)+cotg^2(x)=using (3) we have
=(-1-cotg^2(x))cotg^2(x)+(cotg^2(x))=
-cotg^2(x)-cotg^4(x)+cotg^2(x)=
-cotg^4(x)
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