can anyone solve this one trigometrical ratio sum--sec A(1-sin A)(sec A+tan A)=1,plz solve this for me?

Answer 1

5

Answer 2

This is an identity to prove, not an equation to solve. They're quite different.

Remember sec is 1/cos and tan is sin/cos.
So sec A(1 - sin A)(sec A + tan A) =
(sec A - tan A)(sec A + tan A) =
sec² A - tan² A =
1

Answer 3

You want us to simplify, or solve for A?

Notice that sec A=1/(cos A), and tan A=(sin A)/(cos A)

Now replace all of the sec A, and tan A:

(sec A)(1-sin A)(sec A+tan A)=(1/cos A)(1-sin A)(1/cos A+ sin A/cos A)=(1/cos A)^2(1-sin A)(1+sin A)= (1/cos A)^2(1-(sin A)^2)= (1/cos A)^2(cos A)^2=1. Thus this is true for all A. (therefore this is a type of trig identity.)

Answer 4

sec A(1-sin A)(sec A+tan A)=1
(sec A-tan A)(sec A + tan A)=1
sec² A - tan² A=1
(1-sin² A)/cos² A=1
cos² A/cos² A=1
1=1

Q.E.D.

Answer 5

consider the lhs
multiply sec a inside the brackets
u get
(sec a - tan a)( sec a + tan a)
ie sec ^2 a - tan^2 a
ie 1 + tan^2 a - tan^2 a
ie 1
lhs = rhs
this is not a question on solving its a question on proving the lhs = rhs

if it were a question on solving the answer would be all real numbers....as this isn an identity......which is true for all values

Answer 6

sec A (1- sin A)(sec A + tan A)

= 1/cos A(1-sin A)(sec A + tan A)
= (1/cos A - tan A)(sec A + tan A)
= (sec A - tan A)(sec A + tan A)
= sec^2 A - tan^2 A
= 1