{"type":"video","version":"1.0","width":1920,"height":1080,"title":"Welcome to the world of AI. - Animated Video By L_83_Yash Ghoderao - Mango Animate","description":"Welcome to the world of AI. animation video uploaded by L_83_Yash Ghoderao. Utilize Trigonometric Identities:We are given tan(2A) = cot(A - 18°).We know: cot(x) = 1 \/ tan(x).Substitute and Simplify:Substitute the identity: tan(2A) = 1 \/ tan(A - 18°).Flip both sides: tan(A - 18°) = 1 \/ tan(2A).Double Angle Identity:Use the double angle identity for tangent: tan(2x) = (2 * tan(x)) \/ (1 - tan²(x)).Substitute and Solve:Substitute the double angle identity: tan(A - 18°) = 1 \/ [(2 * tan(A)) \/ (1 - tan²(A))].Multiply both sides by the denominator on the right: tan(A - 18°) * (1 - tan²(A)) = 1.Matching Angles:The equation suggests that tan(A - 18°) and 1 \/ tan(A) have the same value.Looking at the tangent table or considering the unit circle, these conditions are met when:A - 18° = A (meaning 18° = 0°, which isn't true) ORA - 18° = 180° - A (complementary angles have the same tangent value)Solve for A:Consider the second solution (A - 18° = 180° - A):Add A to both sides: 2A = 198°.Since 2A is given as an acute angle (less than 90°), we can divide both sides by 2: A = 99°.Verify Constraint:We found A = 99°.Since 2 * 99° (198°) is indeed greater than 90°, this solution satisfies the constraint of 2A being acute.Therefore, the value of A in this scenario is A = 99°. Make your animation and host online for free!","url":"https:\/\/mangoanimate.com\/w\/wl5epckyjfnckyw\/welcome-to-the-world-of-ai\/wb3epclytfdcq\/","author_name":"L_83_Yash Ghoderao","author_url":"https:\/\/mangoanimate.com\/homepage\/1eedfcca-3eec-62d6-95be-f23c915625cf","provider_name":"Mango Animate","provider_url":"https:\/\/mangoanimate.com","thumbnail_url":"https:\/\/online.mangoanimate.com\/ai\/w\/119237783131890880\/1\/thumb.jpg","thumbnail_width":1920,"thumbnail_height":1080,"html":"<iframe src=\"https:\/\/mangoanimate.com\/w\/wl5epckyjfnckyw\/welcome-to-the-world-of-ai\/wb3epclytfdcq\/?type=embed\" width=\"840px\" height=\"473px\" frameborder=\"0\" scrolling=\"no\" webkitAllowFullScreen mozallowfullscreen allowFullScreen><\/iframe>"}