If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
Chapter 6 – Trigonometry – Text Book Solution
Practice Set 6.1| Q 4 | Page 131
If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
Solution
We know that sec(θ) = 1/cos(θ) and cosec(θ) = 1/sin(θ), so we can rewrite the given equation as:
5/cos(θ) – 12/sin(θ) = 0
Multiplying both sides by cos(θ) * sin(θ), we get:
5sin(θ) – 12cos(θ) = 0
Dividing both sides by 5cos(θ), we get:
sin(θ)/cos(θ) = 12/5
This is the same as tan(θ) = 12/5.
Now we can use the Pythagorean theorem to find the value of the hypotenuse:
sin²(θ) + cos²(θ) = 1
Substituting sin(θ)/cos(θ) = 12/5, we get:
(12/5)² + cos²(θ) = 1
Solving for cos(θ), we get:
cos(θ) = ± √[1 – (12/5)²] = ± √(119)/5
Since 5sec(θ) – 12cosec(θ) = 0, we know that either sec(θ) = 12/5 or cosec(θ) = 5/12.
If sec(θ) = 12/5, then cos(θ) = 5/12, which contradicts the value we just found for cos(θ).
Therefore, cosec(θ) = 5/12 and sin(θ) = 12/5.
Finally, we can use the definition of sec(θ) = 1/cos(θ) to find the value of sec(θ):
sec(θ) = 1/cos(θ) = 1/[(± √(119))/5] = ± 5/√(119)
So, the values of sec(θ), cos(θ), and sin(θ) are:
sec(θ) = ± 5/√(119), cos(θ) = ± √(119)/5, sin(θ) = 12/5.
Chapter 6 – Trigonometry – Text Book Solution
Practice set 6.1 |Q 4 | P 131
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