Question

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Answer 1

Answer:

First, rationalise the denominator then u will get the (a-b)^2 identity in the numerator so the square will be cancelled out from the root and in the deno. u will get the (a^2-b^2) identity, just simplify it. Now, use the trigonometry identity.

Then,just separate the cosA in the fraction to make two terms. Then just convert the fraction into secA and TanA...

Hope this will help you...

Answer 2

1. First multiply the numerator and denominator by (1 - Sin A)

Now in the numerator we will get (1 - Sin A)^2

and in the denominator we get (1 - SIn^2 A) = Cos^2

So, the part inside root becomes [(1 - Sin A)^2/(Cos^2 A)]

And it becomes [(1 - Sin A)/(Cos A)]^2

So the root over [(1 - Sin A)/(Cos A)]^2 = (1 - Sin A)/(Cos A)

              = 1/Cos A - Sin A/Cos A = Sec A - Tan A

   Hence proved.

2. Similarly,

  Here also multiply the numerator and denominator by (1 - Cos A)

You will get, [(1 - Cos A)^2/(1 - Cos^2 A)] = [(1 - Cos A)^2/(Sin^2 A)]

So root over [(1 - Cos A)/(1 + Cos A)]

= root over [(1 - Cos A)^2/(Sin A)^2 = [(1 - Cos A)/(Sin A)]

= 1/Sin A - Cos A/Sin A = Cosec A - Cot A

 Hence proved.