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CBSE Class 10 Answered

Solve:
Asked by shobhitdc | 13 Sep, 2010, 08:34: PM
answered-by-expert Expert Answer
(1 - cotA + cosecA) (1 + tanA - secA) = 2
(1 - cosA/sinA + 1/sinA) (1 + sinA/cosA - 1/cosA) = 2
[(sinA - cosA + 1)(cosA + sinA - 1)]/(sinAcosA) = 2
[-2cosA(cosA - 1)]/(sinAcosA) = 2                 ... multiplying in the numerator
(cosA - 1) = -sinA     
(cosA - 1)2 = -sin2A        ... squaring both sides,
cos2A - 2cosA + 1 = cos2A -1   
cos A = 1
Hence A = 0, 2π, 4π, ...
Regards,
Team,
TopperLearning.
Answered by | 13 Sep, 2010, 09:34: PM

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