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Popular Trigonometry >

prove 1+1/(cos(x))=(tan^2(x))/(sec(x)-1)

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Solution

prove 1+cos(x)1​=sec(x)−1tan2(x)​

Solution

True
Solution steps
1+cos(x)1​=sec(x)−1tan2(x)​
Manipulating right sidesec(x)−1tan2(x)​
Rewrite using trig identities
sec(x)−1tan2(x)​
Use the Pythagorean identity: tan2(x)+1=sec2(x)tan2(x)=sec2(x)−1=sec(x)−1sec2(x)−1​
Simplify sec(x)−1sec2(x)−1​:sec(x)+1
sec(x)−1sec2(x)−1​
Factor sec2(x)−1:(sec(x)+1)(sec(x)−1)
sec2(x)−1
Rewrite 1 as 12=sec2(x)−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)sec2(x)−12=(sec(x)+1)(sec(x)−1)=(sec(x)+1)(sec(x)−1)
=sec(x)−1(sec(x)+1)(sec(x)−1)​
Cancel the common factor: sec(x)−1=sec(x)+1
=sec(x)+1
=sec(x)+1
Express with sin, cos
1+sec(x)
Use the basic trigonometric identity: sec(x)=cos(x)1​=1+cos(x)1​
Simplify 1+cos(x)1​:cos(x)cos(x)+1​
1+cos(x)1​
Convert element to fraction: 1=cos(x)1cos(x)​=cos(x)1⋅cos(x)​+cos(x)1​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=cos(x)1⋅cos(x)+1​
Multiply: 1⋅cos(x)=cos(x)=cos(x)cos(x)+1​
=cos(x)cos(x)+1​
=cos(x)1+cos(x)​
Expand cos(x)1+cos(x)​:cos(x)1​+1
cos(x)1+cos(x)​
Apply the fraction rule: ca±b​=ca​±cb​cos(x)1+cos(x)​=cos(x)1​+cos(x)cos(x)​=cos(x)1​+cos(x)cos(x)​
Apply rule aa​=1cos(x)cos(x)​=1=cos(x)1​+1
=1+cos(x)1​
We showed that the two sides could take the same form⇒True

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Frequently Asked Questions (FAQ)

  • Is 1+1/(cos(x))=(tan^2(x))/(sec(x)-1) ?

    The answer to whether 1+1/(cos(x))=(tan^2(x))/(sec(x)-1) is True
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