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Identity Tan(squared)x+1=Sec(squared)x
Main formulas:
- For any angle x for which the tangent and secant are defined, we have tan2 x + 1 = sec2 x.
- For any angle x for which the tangent and secant are defined, we have cot2 x + 1 = csc2 x.
Example 1:
Prove the identity tan2 x + 1 = sec2 x.Example 2:
Given that tanθ = − 4.21 and (π/2) < θ < π , find secθ .Example 3:
Prove the following trigonometric identity:
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Example 4:
Prove the identity cot2 x + 1 = csc2 x.Example 5:
Given that secθ = (13/12) and 270° < θ < 360° , find tanθ .Identity Tan(squared)x+1=Sec(squared)x
Find arccos [ cos( − [(π)/3]) ]
- Plot the point - [(π)/3] on the unit circle. Note: − [(π)/3] = − 60°
- Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of − [(π)/3] between 0 and π because 0 ≤ arccos ≤ π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π
arccos[ cos( − [(π)/3]) ] = [(π)/3]
Find arcsin[ sin( − [(4π)/3]) ]
- Plot the point - [(4π)/3] on the unit circle. Note: − [(4π)/3] = − 240°
- Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of − [(4π)/3] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]
arcsin[ sin( − [(4π)/3]) ] = − [(π)/3]
Find arctan[ tan([(5π)/4]) ]
- Plot the point [(5π)/4] on the unit circle. Note: [(5π)/4] = 225°
- Recall that sine is the y value of the point you just plotted. Now find the angle that has the same tangent value of [(5π)/4] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]
arctan[ tan([(5π)/4]) ] = [(π)/4]
Find arccos[ cos([(5π)/4]) ]
- Plot the point [(5π)/4] on the unit circle. Note: [(5π)/4] = 225°
- Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of [5p/4] between 0 and π because 0 ≤ arccos ≤ π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π
arccos[ cos([(5π)/4]) ] = [(3π)/4]
Find arctan[ tan( − [(5π)/6]) ]
- Plot the point − [(5π)/6] on the unit circle. Note: − [(5π)/6] = − 150°
- Recall that tangent is [x/y]. Now find the angle that has the same tangent value of − [(5π)/6] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]
arctan[ tan( − [(5π)/6]) ] = [(π)/6]
Find arcsin[ sin( − [(3π)/4]) ]
- Plot the point − [(3π)/4] on the unit circle. Note: − [(3π)/4] = − 135°
- Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of − [(3π)/4] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]
arcsin[ sin( − [(3π)/4]) ] = − [(π)/4]
Find arccos[ cos([(7π)/6]) ]
- Plot the point [(7π)/6] on the unit circle. Note: [(7π)/6] = 210°
- Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of [(7π)/6] between 0 and π because 0 ≤ arccos ≤ π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π
arccos[ cos([(7π)/6]) ] = [(5π)/6]
Find arcsin[ sin([(4π)/3]) ]
- Plot the point [(4π)/3] on the unit circle. Note: [(4π)/3] = 240°
- Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of [(4π)/3] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]
arcsin[ sin([(4π)/3]) ] = − [(π)/3]
Find arctan[ tan([(3π)/4]) ]
- Plot the point [(3π)/4] on the unit circle. Note: [3p/4] = 135°
- Recall that tangent is [x/y]. Now find the angle that has the same tangent value of [(3π)/4] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]
arctan[ tan([(3π)/4]) ] = − [(π)/6]
Find arccos[ cos( − [(π)/3]) ]
- Plot the point − [(π)/3] on the unit circle. Note: − [(π)/3] = − 60°
- Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of − [(π)/3] between 0 and π because 0 ≤ arccos ≤ π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π
arccos[ cos( − [(π)/3]) ] = [(π)/3]
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Identity Tan(squared)x+1=Sec(squared)x
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
































1 answer
Sun May 12, 2013 5:30 PM
Post by Hero Miles on May 16, 2011
I converted example III into sines and cosines first. It was much easier.
1 answer
Sun May 12, 2013 5:33 PM
Post by Jonathan Taylor on October 26, 2011
Ex.3 IS JUST CONFUSING IS THERE A SIMPLER FORM
1 answer
Sun May 12, 2013 5:40 PM
Post by William Davis on December 28, 2011
In example III he's using strategies I ever knew you could do, which i believe makes solving complex identities treacherous. You should be able to still solve it without looking for shortcuts. but when I try, I am unable to. It's very frustrating.
1 answer
Sat Jun 8, 2013 6:01 PM
Post by William Davis on June 6 at 11:35:47 PM
You are by far the best lecturer this site has. You actually care and put forth effort unlike a lot of profs on this site. I wish you taught the Cal I section.