Prove that
tan2A = (sec2A+1)√sec²A-1
Tan2a=(sev2a+1)√(sec2a-1)
To prove the given expression, we need to work with the left-hand side (LHS) and the right-hand side (RHS) separately and show that they are equal.
Starting with the LHS:
tan(2A)
We can rewrite the tangent of a double angle using the identity:
tan(2A) = 2tan(A) / (1 - tan²(A))
Next, we consider the RHS:
(sec²A + 1)√(sec²A - 1)
We can start by simplifying the expression under the square root:
(sec²A - 1) = (sec²A - 1²) = (sec²A - tan²A)
Simplifying further, we use the identity:
sec²A - tan²A = 1
Now we can substitute this value back into the RHS expression:
(sec²A + 1)√(sec²A - 1)
= (sec²A + 1)√1
= (sec²A + 1)
Since tan(2A) simplifies to (2tan(A) / (1 - tan²(A))) and (sec²A + 1)√(sec²A - 1) simplifies to (sec²A + 1), we can equate the two expressions:
2tan(A) / (1 - tan²(A)) = (sec²A + 1)
To proceed further, let's manipulate the left-hand side of the equation:
2tan(A) / (1 - tan²(A))
= 2tan(A) / sec²(A)
= 2sin(A) / cos(A) * cos²(A) / (1 - sin²(A) / cos²(A))
= 2sin(A) * cos(A) / (cos²(A) - sin²(A))
= 2sin(A) * cos(A) / cos²(A) - sin²(A)
= 2sin(A) * cos(A) / cos²(A) - (1 - cos²(A))
= 2sin(A) * cos(A) / cos²(A) - 1 + cos²(A)
= 2sin(A) * cos(A) - cos²(A) + cos²(A) / cos²(A)
= 2sin(A) * cos(A) / cos²(A)
= 2sin(A) / cos(A)
= 2tan(A)
Now we see that the LHS simplifies to 2tan(A), which is equal to the RHS (sec²A + 1). Therefore, we have proven that:
tan(2A) = (sec²A + 1)√(sec²A - 1)
Tan2A=(sec2A+1)√(sec²A-1)
As has been pointed out to you before ....
You have posted about 20 of these rather complicated trig identities in the last week or so.
Not once have you shown a single step of your own effort, and not once have you even acknowledged the efforts of tutors to give you detailed and lengthy answers.
We can't even tell if you even look at the answers.