Grouping the terms inside the square root gives tan² A + 2 + cot² A. Since tan A cot A = 1, we can replace 2 with 2tan Acot A to form a perfect square.

Recognizing the algebraic expansion a² + 2ab + b² = (a + b)², the expression becomes √((tan A + cot A)²). Taking the square root yields tan A + cot A, which matches the right-hand side.

Prove that $\sqrt{\sec^2 A + \csc^2 A} = \tan A + \cot A$

Answer: Hence proved, $\sqrt{\sec^2 A + \operatorname{cosec}^2 A} = \tan A + \cot A$ .

Step-by-step solution

Step 1: Consider LHS and substitute fundamental identities

We start with the left-hand side of the identity. We substitute the standard trigonometric identities $\sec^2 A = 1 + \tan^2 A$ and $\operatorname{cosec}^2 A = 1 + \cot^2 A$ .

Step 2: Rearrange and rewrite the constant term

Grouping the terms inside the square root gives $\tan^2 A + 2 + \cot^2 A$ . Since $\tan A \cot A = 1$ , we can replace $2$ with $2\tan A\cot A$ to form a perfect square.

Step 3: Factor as a perfect square and simplify

Recognizing the algebraic expansion $a^2 + 2ab + b^2 = (a + b)^2$ , the expression becomes $\sqrt{(\tan A + \cot A)^2}$ . Taking the square root yields $\tan A + \cot A$ , which matches the right-hand side.

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Prove that sec⁡2A+csc⁡2A=tan⁡A+cot⁡A\sqrt{\sec^2 A + \csc^2 A} = \tan A + \cot Asec2A+csc2A​=tanA+cotA

Step-by-step solution

Step 1: Consider LHS and substitute fundamental identities

Step 2: Rearrange and rewrite the constant term

Step 3: Factor as a perfect square and simplify

Prove that $\sqrt{\sec^2 A + \csc^2 A} = \tan A + \cot A$