Math, asked by rajeev7, 1 year ago

cotA+sinA/cotA-sinA=1-cosA+secA/1+cosA-secA

Answers

Answered by qais

11

RHS
(1-cosA+secA)/(1+cosA- secA)

multiplying numerator and denominator by cosA,

(cosA- cos²A +1)/(cosA+ cos²A- 1)

∵1- cos²A = sin²A
∴(cosA- cos²A +1)/(cosA+ cos²A- 1)
=(cosA +sin²A)/(cosA - sin²A)

dividing numerator and denominator by sinA,

(cotA +sinA)/(cotA-sinA)
= LHS

Answered by reachemmanuelm

4

RHS

(1-cosA+secA)/(1+cosA- secA)

multiplying numerator and denominator by cosA,

(cosA- cos²A +1)/(cosA+ cos²A- 1)

∵1- cos²A = sin²A

∴(cosA- cos²A +1)/(cosA+ cos²A- 1)

=(cosA +sin²A)/(cosA - sin²A)

dividing numerator and denominator by sinA,

(cotA +sinA)/(cotA-sinA)

= LHS

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