prove: (tan x + sec x)^2 = 2sec^2 x + 2tan x sec x - 1
expand left side
recall that tan^2 = sec^2 - 1
done
left side
(sin/cos + 1/cos)^2
(1/cos^2)(sin^2 + 2 sin + 1)
right side
2/cos^2 + 2 sin/cos^2 - cos^2/cos^2)
(1/cos^2)( 2 + 2 sin -(1-sin^2)
(1/cos^2)(2 + 2 sin -1 + sin^2)
(1/cos^2)( sin^2 + 2 sin + 1) remarkable
Much more fun to do it the hard way
To prove the given equation, we'll start with the left-hand side (LHS) and try to manipulate it step by step until we reach the right-hand side (RHS). Here's how we can do it:
LHS: (tan x + sec x)^2
Step 1: Expand the square
LHS = (tan x + sec x)(tan x + sec x)
Step 2: Apply the distributive property
LHS = tan x(tan x) + tan x(sec x) + sec x(tan x) + sec x(sec x)
Step 3: Simplify each term
LHS = tan^2 x + tan x sec x + tan x sec x + sec^2 x
Step 4: Combine like terms
LHS = tan^2 x + 2tan x sec x + sec^2 x
Now, let's compare this expression with the RHS of the equation:
RHS: 2sec^2 x + 2tan x sec x - 1
To prove that the LHS is equal to the RHS, we need to show that both sides simplify to the same expression.
Step 5: Simplify the RHS
RHS = 2sec^2 x + 2tan x sec x - 1
Step 6: Combine like terms
RHS = 2sec^2 x + 2tan x sec x - 1
Notice that the simplified expression on the RHS is the same as the expression we obtained from the LHS. Therefore, we have successfully proven that:
(tan x + sec x)^2 = 2sec^2 x + 2tan x sec x - 1