98.7k views
5 votes
Prove 1-sin 2A/1+sin2A=(cotA-1/cotA+1)^2​

User Probablyup
by
9.2k points

1 Answer

1 vote

Answer:

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.

User Agrawal Shraddha
by
7.7k points

Related questions

asked Dec 8, 2019 215k views
Mario Aguilera asked Dec 8, 2019
by Mario Aguilera
7.8k points
1 answer
2 votes
215k views
asked Aug 2, 2021 190k views
Ckuri asked Aug 2, 2021
by Ckuri
8.9k points
1 answer
4 votes
190k views
asked Apr 2, 2021 181k views
Paril asked Apr 2, 2021
by Paril
8.4k points
1 answer
0 votes
181k views