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prove 1-sin 2A/1+sin2A=(cotA-1/cotA+1)^2​

Sagot :

Answer:

Step-by-step explanation:

Take the right side of the given expression:

(cotA - 1/ cotA + 1)^2​

= (cot A - 1)^2  / (cotA + 1)^2

= (cot^2A + 1 - 2 cotA) / ( (cot^2A + 1 + 2 cotA)

Now cosec^2 A = 1 + cot^2A   so we have:

= (cosec^2A - 2 cotA) / (cosec^2A + 2 cot A)  

=( 1/ sin^2A) - 2cosA/sinA

  --------------------------------

  (1/ sin^2A ) + 2cosA/sinA

=   (1 - 2 sinAcosA) / sin^2A

    ----------------------------------

    (1 +2 sinAcosA) / sin^2A

=   (1 - 2 sinAcosA)  /  (1 +2 sinAcosA)

Now 2 sinAcosA = sin2A so this simplifies to:

(1 - sin2A / (1 + sin2A).

Which is the left side of the original identity.

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