The expression cose c 2 A cot 2 A− sec 2 A tan 2 A−( cot 2 A− tan 2 A)( sec 2 A+cose c 2 A−1) is equal to

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The expression $cos e c 2 A cot 2 A − sec 2 A tan 2 A − (cot 2 A − tan 2 A) (sec 2 A + cos e c 2 A − 1)$ is equal to

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sec 2 A

cos e c 2 A

answer is A.

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Detailed Solution

First, we write required trigonometric properties as shown below,

⇒ sin 2 x + cos 2 x = 1

⇒ tan x = sin x cos x

⇒ sec x = 1 cos x

Now, we write the required algebraic formulas as,

⇒ x 3 − y 3 = (x − y) (x 2 + y 2 + x y)

⇒ x 2 − y 2 = (x + y) 2 − 2 x y

In this question, we have given trigonometric expression as,

cos e c 2 A cot 2 A − sec 2 A tan 2 A − (cot 2 A − tan 2 A) (sec 2 A + cos e c 2 A − 1)

Now we will break the given trigonometric expression into two parts or the trigonometric expression can be written as P − Q.
Here,

P = cos e c 2 A cot 2 A − sec 2 A tan 2 A and = (cot 2 A − tan 2 A) (sec 2 A + cos e c 2 A − 1)

Now we can write the expression as,

⇒ cos e c 2 A cot 2 A − sec 2 A tan 2 A − (cot 2 A − tan 2 A) (sec 2 A + cos e c 2 A − 1) = P − Q

First, we will solve the term P by expanding the

cos e c

and

cot

terms in

sin

and

cos

terms because it is easy to solve and calculate.

P = cos e c 2 A cot 2 A − sec 2 A tan 2 A

Now we will apply the trigonometric properties as,

⇒ P = 1 sin 2 A × cos 2 A sin 2 A − 1 cos 2 A × sin 2 A cos 2 A

Now, we will simplify the above expression as,

⇒ P = cos 2 A sin 4 A − sin 2 A cos 4 A

After simplification we will get,

⇒ P = cos 6 A − sin 6 A sin 4 A cos 4 A

Now we will use the algebraic formula in the numerator of the expression to find the easiest form and then solve it accordingly.

⇒ P = cos 2 A 3 − sin 2 A 3 sin 4 A cos 4 A

Now, we will apply the algebraic properties as,

⇒ P = cos 2 A − sin 2 A cos 4 A + sin 4 A + sin 2 A cos 2 A sin 4 A cos 4 A

Now, we will simplify the above expression as,

⇒ P = cos 2 A − sin 2 A cos 2 A + sin 2 A 2 − sin 2 A cos 2 A sin 4 A cos 4 A

After simplification we will get,
First, we will solve the term Q by expanding the

cosec

and

cot

terms in

sin

and

cos

terms because it is easy to solve and calculate.

⇒ Q = 1 sin 2 A cos 2 A sin 2 A − sin 2 A cos 2 A 1 cos 2 A + 1 sin 2 A − 1

Now, we will apply the algebraic properties as,

⇒ Q = cos 4 A − sin 4 A sin 2 A cos 2 A 1 − sin 2 A cos 2 A sin 2 A cos 2 A

After simplification we will get,

⇒ Q = cos 2 A − sin 2 A 1 − sin 2 A cos 2 A sin 4 A cos 4 A

Now, we will substitute

cos 2 A − sin 2 A 1 − sin 2 A cos 2 A sin 4 A cos 4 A

for P and

cos 2 A − sin 2 A 1 − sin 2 A cos 2 A sin 4 A cos 4 A

for Q in the expression

P − Q

to find the solution of the trigonometric expression.

⇒ P − Q = cos 2 A − sin 2 A 1 − sin 2 A cos 2 A sin 4 A cos 4 A − cos 2 A − sin 2 A 1 − sin 2 A cos 2 A sin 4 A cos 4 A

After simplification, we will get

∴ P − Q = 0

Therefore, the solution of the trigonometric expression

cos e c 2 A cot 2 A − sec 2 A tan 2 A − (cot 2 A − tan 2 A) (sec 2 A + cos e c 2 A − 1)

is 0.

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