We have proven that the trigonometric identity [(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] equals 1 + (secθ * cosec θ)
<h3>How to solve Trigonometric Identities?</h3>
We want to prove the trigonometric identity;
[(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] = 1 + sec θ
The left hand side can be expressed as;
[(tan θ)/(1 - (1/tan θ)] + [(1/tan θ)/(1 - tan θ)]
⇒ [tan²θ/(tanθ - 1)] - [1/(tan θ(tanθ - 1)]
Taking the LCM and multiplying gives;
(tan³θ - 1)/(tanθ(tanθ - 1))
This can also be expressed as;
(tan³θ - 1³)/(tanθ(tanθ - 1))
By expansion of algebra this gives;
[(tanθ - 1)(tan²θ + tanθ.1 + 1²)]/[tanθ(tanθ(tanθ - 1))]
Solving Further gives;
(sec²θ + tanθ)/tanθ
⇒ sec²θ * cotθ + 1
⇒ (1/cos²θ * cos θ/sin θ) + 1
⇒ (1/cos θ * 1/sin θ) + 1
⇒ 1 + (secθ * cosec θ)
Read more about Trigonometric Identities at; brainly.com/question/7331447
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Answer:
x = - 6
Step-by-step explanation:
Let x b number, according to instructions´ we must
1.-Multiply by 4
2.-Subtract 6
3.-Divide by 2
4.-Arturo says his new number - 15
4,.Which one was the original number ??
The maths expression for that is :
(4*x - 6 ) / 2 = - 15
we solve for x
4x - 6 = - 30
4x = - 24
x = - 6
The factors of rty are
1, r, t, y,
rt, ry, ty, rty
Answer:
C, 20 units
Step-by-step explanation:
We see that both angles QRS and QTR are 90 degrees. In addition, angles SQR and RQT are equivalent (because they're both angle Q).
By AA Similarity, we know that triangle QTR is similar to triangle QRS.
With this similarity in mind, we can look at the ratios of corresponding lengths to set up a proportion. QR from triangle QTR is the hypotenuse, and it corresponds to hypotenuse QS from triangle QRS. So, we can write the ratio x/(9 + 16) = x/25.
Now, we see that long leg QT of triangle QTR corresponds to long leg QR of triangle QRS. So, another ratio we can write is: 16/x.
Finally, we set these two ratios equal to each other:
Cross-multiplying, we get: 
.
Thus, x = 
. The answer is C, 20 units.
Hope this helps!
Answer: - 0.28
Explanation:
1) Expected value: is the weighted average of the values, being the probabilities the weight.
That is: ∑ of prbability of event i × value of event i.
In this case: (probability of getting 2 or 12) × (+6) + (probability of gettin 3 or 11) × (+2) + (probability of any other sum) × (-1).
2) Sample space:
Sum Points awarded
1+ 1 = 2 +6
1 + 2 = 3 +2
1 + 3 = 4 -1
1 + 4 = 5 -1
1 + 5 = 6 -1
1 + 6 = 7 -1
2 + 1 = 3 +2
2 + 2 = 4 -1
2 + 3 = 5 -1
2 + 4 = 6 -1
2 + 5 = 7 -1
2 + 6 = 8 -1
3 + 1 = 4 -1
3 + 2 = 5 -1
3 + 3 = 6 -1
3 + 4 = 7 -1
3 + 5 = 8 -1
3 + 6 = 9 -1
4 + 1 = 5 -1
4 + 2 = 6 -1
4 + 3 = 7 -1
4 + 4 = 8 -1
4 + 5 = 9 -1
4 + 6 = 10 -1
5 + 1 = 6 -1
5 + 2 = 7 -1
5 + 3 = 8 -1
5 + 4 = 9 -1
5 + 5 = 10 -1
5 + 6 = 11 +2
6 + 1 = 7 -1
6 + 2 = 8 -1
6 + 3 = 9 -1
6 + 4 = 10 -1
6 + 5 = 11 +2
6 + 6 = 12 +6
2) Probabilities
From that, there is:
- 2/36 probabilities to earn + 6 points.
- 4/36 probabilites to earn + 2 points
- the rest, 30/36 probabilities to earn - 1 points
3) Expected value = (2/36)(+6) + (4/36) (+2) + (30/36) (-1) = - 0.28
