Here is a neat trick:
to see if your equation is a true identity, graph the left side and the right side as
separate functions using something like Wolfram. If the two curves coincide, then
you have an identity.
e.g. prove that cos^2 x - sin^2 x = 2cos^2 x - 1
https://www.wolframalpha.com/input/?i=graph+y+%3D+cos%5E2+x+-+sin%5E2+x+%2C+y+%3D+2cos%5E2+x+-+1
in our case:
https://www.wolframalpha.com/input/?i=graph+y+%3D+%28tan%5E2x%29%2F%28cos%5E2x%29%2Bsec%5E2x%2Bcsc%5E2x%2C+y%3Dsec%5E4x%2Bcsc%5E2x*cos%5E2x%2B1
Another way is to pick a standard angle, e.g. 30° and see if the equation is true.
If it is false, then of course you can stop, since all you need is one exception for
an equation NOT to be an identity. If it is true, there is a high probability that you
have an identity
it looks like we have an identity.
A common method is to take the left side and the right side and independently simplify
until you end up with the same expression for LS and RS.
Unless you recognize one of the standard relations, change all expressions to sines and cosines
and hope for the best
RS = sec^4x + csc^2x*cos^2x + 1
LS = (tan^2x)/(cos^2x)+sec^2x+csc^2x
= (sec^2 x - 1)/cos^2 x + sec^2 x + csc^2 x , (since tan^2 x + 1 = sec^2 x)
= 1/cos^4 x - sec^2 x + sec^2 x + csc^2 x
= sec^4 + csc^2 x
jump to RS
RS = sec^4x + csc^2x*cos^2x + 1
= sec^4 x + (1/sin^2 x)(cos^2 x) + 1
= sec^4 x + cot^2 x + 1
= sec^4 x + (csc^2 x -1) + 1
= sec^4 x + csc^2 x
= LS
try the others in a similar way
Equation 1: (tan^2x)/(cos^2x)+sec^2x+csc^2x=sec^4x+csc^2x*cos^2x+1
Equation 2: (csc^2x*(sin^4x+cos^2x))/(cos^2x)-cos^2x = tan^2x*csc^2x+cot^2x+sin^2x-1
Equation 3: tan^2x*sin^2x - (cot^2x)/(csc^2x) = -(sin^2x*cot^2x)/(csc^2x)- cos^4x
Do the following for each equation:
a. Verify if each equation is a trig identity.
b. Explain why or why not.
1 answer
1 answer