Question

1-cos2A= ????????? and 1% ¶÷3348 ©®​

Answer 1

Answer:

Since Cos(A+B)= CosA.CosB - SinA.SinB ( trigonometric identity)

=> Cos( A+A) = Cos2A = Cos²A - Sin²A

=> Cos²A -( 1-Cos²A)

=> 2Cos²A -1

=> 2(1-Sin²A)-1

=> 2 - 2Sin²A -1

=> 1–2Sin²A

=> 1- Sin²A -Sin²A

=> Cos²A - Sin²A

=> (Cos²A - Sin²A) / ( Cos²A + Sin²A) Since Cos²A + Sin²A = 1

Now, by dividing numerator & denominator by Cos²A

We get, Cos2A =( 1- tan²A) / (1+ tan²A) ………(1)

NOW, by putting up the above value in

(1- Cos2A) / (1+Cos2A)

1 -Cos2A = 1 - ( 1-tan²A) /(1+tan²A)

= (1+tan²A - 1 + tan²A)

= 2tan²A / (1+tan²A)….(2)

Now similarly, 1+ Cos2A = 2/(1+ tan²A) …….(3)

Finally, (1-Cos2A) / (1+Cos2A)

= [2tan²A / (1+tan²A)] * [(1+tan²A) / 2 ]

= tan²A

This is probably your answer.

Hope this helps.

Explanation:

Answer 2

Answer:

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Explanation:

sorry can you please repate the question