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Prove that ABBAABBAsin(4A-2B)+sin(4B-2A)cos(4A-2B)+cos(4B-2A) = tan(A + B)

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प्रश्न

Prove that `(sin(4"A" - 2"B") + sin(4"B" - 2"A"))/(cos(4"A" - 2"B") + cos(4"B" - 2"A"))` = tan(A + B)

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उत्तर

`(sin(4"A" - 2"B") + sin(4"B" - 2"A"))/(cos(4"A" - 2"B") + cos(4"B" - 2"A")) = (sin{(4"A" - 2"B" + 4"B" - 2"A")/2} cos {(4"A" - 2"B" - 4"B" + 2"A")/2})/(cos{(4"A" - 2"B" + 4"B" - 2"A")/2} cos{(4"A" - 2"B" - 4"B" + 2"A")/2})`

= `(sin((2"A" + 2"B")/2) * cos((6"A" - 6"B")/2))/(cos((2"A" + 2"B")/2) * cos((6"A" - 6"B")/2)`

= `(sin("A" + "B"))/(cos("A" + "B"))`

= tan(A + B)

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Trigonometric Functions and Their Properties
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.6 [पृष्ठ १२२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.6 | Q 13 | पृष्ठ १२२

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