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Popular Trigonometry >

sec(θ)csc(θ)=1

  • Pre Algebra
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Solution

sec(θ)csc(θ)=1

Solution

NoSolutionforθ∈R
Solution steps
sec(θ)csc(θ)=1
Subtract 1 from both sidessec(θ)csc(θ)−1=0
Express with sin, cos
−1+csc(θ)sec(θ)
Use the basic trigonometric identity: csc(x)=sin(x)1​=−1+sin(θ)1​sec(θ)
Use the basic trigonometric identity: sec(x)=cos(x)1​=−1+sin(θ)1​⋅cos(θ)1​
Simplify −1+sin(θ)1​⋅cos(θ)1​:sin(θ)cos(θ)−sin(θ)cos(θ)+1​
−1+sin(θ)1​⋅cos(θ)1​
sin(θ)1​⋅cos(θ)1​=sin(θ)cos(θ)1​
sin(θ)1​⋅cos(θ)1​
Multiply fractions: ba​⋅dc​=b⋅da⋅c​=sin(θ)cos(θ)1⋅1​
Multiply the numbers: 1⋅1=1=sin(θ)cos(θ)1​
=−1+sin(θ)cos(θ)1​
Convert element to fraction: 1=sin(θ)cos(θ)1sin(θ)cos(θ)​=−sin(θ)cos(θ)1⋅sin(θ)cos(θ)​+sin(θ)cos(θ)1​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=sin(θ)cos(θ)−1⋅sin(θ)cos(θ)+1​
Multiply: 1⋅sin(θ)=sin(θ)=sin(θ)cos(θ)−sin(θ)cos(θ)+1​
=sin(θ)cos(θ)−sin(θ)cos(θ)+1​
cos(θ)sin(θ)1−cos(θ)sin(θ)​=0
g(x)f(x)​=0⇒f(x)=01−cos(θ)sin(θ)=0
Rewrite using trig identities
1−cos(θ)sin(θ)
Use the Double Angle identity: 2sin(x)cos(x)=sin(2x)sin(x)cos(x)=2sin(2x)​=1−2sin(2θ)​
1−2sin(2θ)​=0
Move 1to the right side
1−2sin(2θ)​=0
Subtract 1 from both sides1−2sin(2θ)​−1=0−1
Simplify−2sin(2θ)​=−1
−2sin(2θ)​=−1
Multiply both sides by 2
−2sin(2θ)​=−1
Multiply both sides by 22(−2sin(2θ)​)=2(−1)
Simplify−sin(2θ)=−2
−sin(2θ)=−2
Divide both sides by −1
−sin(2θ)=−2
Divide both sides by −1−1−sin(2θ)​=−1−2​
Simplifysin(2θ)=2
sin(2θ)=2
−1≤sin(x)≤1NoSolutionforθ∈R

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Popular Examples

prove cos(a+270)=sin(a)1+sin(θ)=2-sin(θ)-4cos^2(θ)+cos(θ)=9cos(θ)+33+4cot(x)=5csc(x)(tan(x)-1)(sec(x)+1)=0

Frequently Asked Questions (FAQ)

  • What is the general solution for sec(θ)csc(θ)=1 ?

    The general solution for sec(θ)csc(θ)=1 is No Solution for θ\in\mathbb{R}
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