Math, asked by granthoundayil, 5 months ago

1/cosecA-cotA-1/sinA=1/sinA-1/cosecA-cotA​

Answers

Answered by keshavjindal868
0

Answer:

hope it will help you

Step-by-step explanation:

(1/cosecA-cotA)-(1/sinA)

={1/(1/sinA-cosA/sinA)}-(1/sinA)

=[1/{(1-cosA)/sinA}]-(1/sinA)

=sinA/(1-cosA)-(1/sinA)

=(sin²A-1+cosA)/sinA(1-cosA)

={(1-cos²A)-(1-cosA)}/sinA(1-cosA)

={(1+cosA)(1-cosA)-(1-cosA)}/sinA(1-cosA)

=(1-cosA)(1+cosA-1)/sinA(1-cosA)

=cosA/sinA

=cotA

RHS

(1/sinA)-(1/cosecA+cotA)

=(1/sinA)-{1/(1/sinA+cosA/sinA)}

=(1/sinA)-1/{(1+cosA)/sinA}

=(1/sinA)-sinA/(1+cosA)

=(1+cosA-sin²A)/sinA(1+cosA)

={1+cosA-(1-cos²A)}/{sinA(1+cosA)}

={(1+cosA)-(1+cosA)(1-cosA)}/{sinA(1+cosA)}

=(1+cosA)(1-1+cosA)/sinA(1+cosA)

=cosA/sinA

=cotA

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