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1+tan A + cot A = 1 + sec A cosec A​

User Bcf Ant
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Answer:

See below for proof.

Explanation:

Given trigonometric identity:


1 + \tan A + \cot A = 1 + \sec A \csc A

To prove the given trigonometric identity, manipulate the left side to match the right side.

Express tan(A) and cot(A) in terms of sine and cosine using the identities tan(A) = sin(A)/cos(A) and cot(A) = cos(A)/sin(A):


1 + (\sin A)/(\cos A) + (\cos A)/(\sin A) = 1 + \sec A \csc A

Combine the fractions on the left side of the equation:


1 + (\sin^2 A)/(\sin A\cos A) + (\cos^2 A)/(\sin A \cos A) = 1 + \sec A \csc A


1 + (\sin^2 A+\cos^2 A)/(\sin A \cos A) = 1 + \sec A \csc A

Apply the Pythagorean trigonometric identity, sin²(A) + cos²(A) = 1:


1 + (1)/(\sin A \cos A) = 1 + \sec A \csc A

Separate the fractions:


1 + (1)/(\sin A)\cdot (1)/(\cos A) = 1 + \sec A \csc A

Use the reciprocal trigonometric identities sec(A)=1/cos(A) and cosec(A)=1/sin(A):


1 + \csc A \sec A = 1 + \sec A \csc A

Apply the commutative property of multiplication:


1 + \sec A \csc A = 1 + \sec A \csc A

Therefore, the left side is equal to the right side, and the trigonometric identity is proven.

User WBLord
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