If cosecθ + cotθ = 5, then evaluate secθ.
Answers
Step-by-step explanation:
secθ = -1 or 1.08(approx)
Step-by-step explanation:
In this question
We have been given that
Cosecθ + Cotθ = 5
Then we need to find the value of Secθ
Changing Cosecθ and Cotθ into Sinθ and Cosθ
Now, 1Sinθ + CosθSinθ = 5\frac{1}{Sin\theta}\ +\ \frac{Cos\theta}{Sin\theta}\ =\ 5
Sinθ
1
+
Sinθ
Cosθ
= 5
1+CosθSinθ = 5\frac{1+Cos\theta}{Sin\theta}\ =\ 5
Sinθ
1+Cosθ
= 5
Cross Multiplying we get,
1 + Cosθ = 5Sinθ
Squaring both the sides we get,
(1 + Cosθ)2 = 25Sin2θ(1\ +\ Cos\theta)^{2}\ =\ 25Sin^{2}\theta(1 + Cosθ)
2
= 25Sin
2
θ
1 + Cos2θ + 2Cosθ = 25Sin2θ1\ +\ Cos^{2}\theta\ +\ 2Cos\theta\ =\ 25Sin^{2}\theta1 + Cos
2
θ + 2Cosθ = 25Sin
2
θ
1 + Cos2θ + 2Cosθ = 25(1−Cos2θ)1\ +\ Cos^{2}\theta\ +\ 2Cos\theta\ =\ 25(1-Cos^{2}\theta)1 + Cos
2
θ + 2Cosθ = 25(1−Cos
2
θ)
1 + Cos2θ + 25Cos2θ + 2Cosθ −25 = 01\ +\ Cos^{2}\theta\ +\ 25Cos^{2}\theta\ +\ 2Cos\theta\ -25\ =\ 01 + Cos
2
θ + 25Cos
2
θ + 2Cosθ −25 = 0
26Cos2θ + 2Cosθ − 24 = 026Cos^{2}\theta\ +\ 2Cos\theta\ -\ 24\ =\ 026Cos
2
θ + 2Cosθ − 24 = 0
Using Shridhanacharya formula we get,
Here a = 26 ; b = 2 ; c = -24
Cosθ = −2 ± (2)2 − 4×26×(−24)2×26\frac{-2\ \pm\ \sqrt{(2)^{2}\ -\ 4\times 26\times (-24)}}{2\times 26}
2×26
−2 ±
(2)
2
− 4×26×(−24)
Cosθ = −2 ± 4 +249652\frac{-2\ \pm\ \sqrt{4\ + 2496}}{52}
52
−2 ±
4 +2496
Cosθ = −2 ± 250052\frac{-2\ \pm\ \sqrt{2500}}{52}
52
−2 ±
2500
Cosθ = −2 ± 5052\frac{-2\ \pm\ 50}{52}
52
−2 ± 50
Cosθ = −2 + 5052\frac{-2\ +\ 50}{52}
52
−2 + 50
and −2 − 5052\frac{-2\ -\ 50}{52}
52
−2 − 50
Cosθ = 4852and[tex]−5252\frac{48}{52} and [tex]\frac{-52}{52}
52
48
and[tex]
52
−52
Therefore we know that Secθ = 1Cosθ\frac{1}{Cos\theta}
Cosθ
1
Hence Secθ = 5248\frac{52}{48}
48
52
and 52−52\frac{52}{-52}
−52
52
Secθ = 1.08 and 1
Answer:
cosecθ+cotθ=5
➡(1+cosθ)/sinθ=5
➡1+cosθ=5sinθ
➡(1+cosθ)²=25sin²θ
➡(1+cosθ)²=25(1-cos²θ)
➡(1+cosθ)²=25(1+cosθ)(1-cosθ)
➡1+cosθ=25(1-cosθ)
➡1+cosθ=25-25cosθ
➡26cosθ=24
➡cosθ=24/26=12/13
➡secθ=13/12