Question

prove each of the following identities:​

Answer 1

Step-by-step explanation:

First write Tan²A as Sin²A/Cos²A

Then you will get Sin²A /Cos²A -Sin²A

then multiply and divide the alone sin²A by Cos²A

then you will get Sin²A-Sin²ACos²A/Cos²A

then take the Sin²A common from numerator

you will get Sin²A(1-Cos²A)/cos²A

then use identity sin²A + cos²A =1 as 1-Cos²A=Sin²A

then write 1-cos²A in numerator as Sin²A

then you will get Sin²A Sin²A/cos²A

then write sin²A/cos²A= Tan²A

then you will get Sin²A Tan²A

hence proved

Answer 2

Step-by-step explanation:

To proof:

(i) tan^2A - sin^2A = tan^2A sin^2A

from LHS

tan^2A -sin^2A

= (sin^2A / cos^2A) - sin^2A.. [tan A=sin A/cos A]

= (sin^2A - sin^2A cos^2A) / cos2A

= sin ^2A (1-cos^2A) / cos^2A [tan A = sinA / cos A]

= tan^2A sin^2A= RHS

(ii)

cot^2A - cos^2A

= cos^2A/sin^2A-cos^2A

= (cos^2A - cos^2A sin^2A)/sin^2A

= cos^2A (1 - sin^2A)/sin^2A

= (cos^2A/ sin^2A) cos^2A

(1- sin^2A)

= cos^2A

= cot^2A cos^2A = RHS

Hope this helps :)