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prove-2sin68/cos22-2tan(90-15)/5cot15=3tan45tan20tan40tan50tan70/5(sin square70+sin square20) |
| Qn. 2sin68/cos22-2tan(90-15)/5cot15=1 + 3tan45tan20tan40tan50tan70/5(sin square70+sin square20) Ans: LHS =[ 2sin68/cos22]-[2tan(90-15)/5cot15] = [2cos22/cos22 ] -[2cot15/5cot15] = 2 - [2/5] = 8/5 RHS = 1 + [3tan45tan20tan40tan50tan70/5(sin square70+sin square20)] = 1 + [3 tan20tan40cot40cot20 / 5(sin270 + cos270] = 1 + 3 / 5 =8/5 LHS=RHS |
