For any angle x for which the tangent and secant are defined, we have tan2x + 1 = sec2x.
For any angle x for which the tangent and secant are defined, we have cot2x + 1 = csc2x.
Prove the identity tan2x + 1 = sec2x.
Given that tanθ
4.21 and (π/2) < θ
, find secθ
Prove the following trigonometric identity:
Prove the identity cot2x + 1 = csc2x.
Given that secθ
= (13/12) and 270°
< θ < 360°
, find tanθ
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The first example here is to prove the identity tan2(x)+1=sec2(x).0102
The trick there is to remember the original Pythagorean identity, which is cos2(x)+sin2(x)=1, that's the original Pythagorean identity.0109
That's one that you really should memorize and remember throughout all your work with trigonometry.0127
What you do to manipulate this into the new identity, is you just divide both sides by cos2(x).0136
On the left, we get cos2(x)/cos2(x) plus, let me write it as (sin(x)/cos(x)2=1/cos2(x).0152
Then of course, cos2(x)/cos2(x) is just 1, sin/cos, remember that's the definition of tangent, so this is tan2(x), 1/cos, that's the definition of sec(x), so sec2(x).0173
You can just rearrange this into tan2(x)+1=sec2(x).0191
That shows that this new identity is just a straight consequence of the original Pythagorean identity, so really if you can remember that Pythagorean identity, you can pretty quickly work out this new Pythagorean identity for tangents and cotangents.0203
Let's try using the Pythagorean identity for tangents and cotangents.0224
In this problem, we're given the tan(θ)=-4.21, and θ is between π/2 and π, we want to find secθ.0230
Remembering the Pythagorean identity, tan2(θ)+1=sec2(θ).0238
If we plug in what we're given here, tan2(θ), that's (-4.212)+1=sec2(θ).0246
Well, 4.212, that's something I'm going to workout on my calculator, is 17.7241.0259
So, 17.7241+1=sec2(θ), that's 18.7241=sec2(θ).0274
If I take the square root of both sides, I get sec(θ) is equal to plus or minus the square root of 18.7241.0290
Again, I'm going to do that square root on the calculator, and I get approximately 4.33.0303
The question here is whether we want the positive or negative square root, and that's where we use the other piece of information given in the problem.0323
Θ is in the second quadrant here, θ is between π/2 and π, so θ is somewhere over here.0332