Question

prove that Sin A + cos a cos square + Cos A + sec inverse x is equal to 7 + tan square A + cot square A​

Answer 1

Answer:

Step-by-step explanation:

Let it be equal to 7+ tan^2 a + cot^ a

Hence proved

Answer 2

Answer:

As we know tan A=Sin A/Cos A and tan B=Sin B/Cos B  So

tan squareA - tan squareB= (Sin A/Cos A)^2 - (Sin B/Cos B)^2 =

{(Sin^2 A Cos^2 B)-(Cos^2 A Sin^2 B)}/(Cos^2 A .Cos^2 B)

As we know that Sin^2 A=1-Cos^2 A and Sin^2 B=1-Cos^2 B ,So

tan squareA - tan squareB={( Cos^2 B(1-cos^2 A))-(Cos^2 A (1-Cos^2 B))}/(Cos^2 A .Cos^2 B)= (Cos^2 B- Cos^2 A)/(Cos^2 A .Cos^2 B)

Step-by-step explanation: