Question

If A + B = 90°, then tanAtanB+tanAcotB/sinAsecB-sin²B/cos²A is equal to
A. cot²A
B. cot² B
C. –tan²A
D. – cot² A

Answer 1

Answer:

a is the correct answer

plz mark as brainliest

Answer 2

C. tan²A

Step-by-step explanation:

Given: A + B = 90°, so B = 90°-A.

So (tanA. tanB + tanA. cotB) / (sinA. secB) - (sin²B/cos²A) =  

= [tan A.tan(90-A) + tan A.Cot (90-A)] /[(sin A. Sec(90-A)] - sin²(90-A)/cos² A

 = tan A.cot A + tan A.tan A /Sin A. Cosec A - cos² a/cos² a

 = 1 + tan² A / 1 - (1)

= 1 + tan² A - 1

= tan² A  

Option C is the answer.

Error in option C. Negative sign shouldn't be there. It is + tan²A