SIR PLEASE PROVE

Asked by gunjansingh | 6th Dec, 2009, 06:18: PM

Expert Answer:

[cot 2A(secA - 1)]/(1+sinA)

Multiply and divide by (1-sinA)(secA + 1) and using x2 - y2 = (x+y)(x-y)

[cot 2A(sec2A - 1)(1-sinA)]/[(1-sin2A)(secA + 1)] =

[cot 2A tan2A(1-sinA)]/[cos2A(secA + 1)] =

[(1-sinA)]/[cos2A(secA + 1)] =

[sec 2A(1-sinA)]/[1+secA]

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Answered by  | 6th Dec, 2009, 07:18: PM

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