Question

If A, B, C are the angles of ΔABC then prove that cosec^2(B+C/2) - tan^2(A/2) = 1

Answer 1

Step-by-step explanation:

In a Δ

sum of all angles=180°

a+b+c=180

    b+c=180-a                                .................1

then in given statement taking LHS

       =cosec²(b+c/2)-tan²(a/2)

 substituting 1 in cosec²(b+c/2)

       =cosec²(180/2-a/2)-tan²(a/2)

       =cosec²(90-a/2)-tan²(a/2)

       (identity  cosec(90-Ф)=secФ)

then =sec²(a/2)-tan²(a/2)

       (identity sec²Ф-tan²Ф=1)

  so  =1 ⇒ RHS 

                                                                                   HENCE PROVED

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Answer 2

Answer:

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