Math, asked by soumya8709, 1 year ago

tan²A-tan²B=cos²B-cos²A/cos²B×cos²A=sin²Asin²B/cos²Acos²B

Answers

Answered by amanpareek33
0

Answer: Given :


tan²A - tan²B = cos²B - cos²A/cos²B cos²A= sin²A- sin²B / cos²B cos²A


LHS = tan²A - tan²B


= sin²A/cos²A - sin²B/cos²B


[tanθ = sinθ /cosθ ]


= (sin²A cos²B - sin²B cos²A)/cos²A cos²B


=[ sin²A(1-sin²B) - sin²B(1-sin²A)]/cos²A cos²B


[cos²θ = 1 - sin²θ ]


=[ (sin²A - sin²Asin²B) - (sin²B - sin²Asin²B)]/cos²A cos²B


= [(sin²A - sin²Asin²B - sin²B + sin²Asin²B)]/cos²A cos²B


=[ (sin²A - sin²B - sin²Asin²B + sin²Asin²B)] /cos²A cos²B


LHS = (sin²A - sin²B)/cos²A cos²B = RHS


Read more on Brainly.in - https://brainly.in/question/5078029#readmore




soumya8709: tysm
Answered by neelamsanjayagarwal1
0
done!!!!
good sum!!!!!
Attachments:
Similar questions