I have another question! I tried multiple times to prove this identity, but still stuck with the last steps. Can you please help me? Thank you so much :)
Explanations (2)
Student Explanation
February 12, 2025
Hi AdorableApatosaurus7671 😁
Thanks for your question!
In this school perseverance week, I'd like to congratulate you on your efforts at school, it's not always easy, but your perseverance makes all the difference 🎉💫. Keep up the good work!
Let's start with the following equivalence.
And, let's complete the resolution.
The final result :
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Student Explanation
February 12, 2025
Salut!
Thank you for using our services! In this school perseverance week, I want to congratulate you for your efforts at school, it is not always easy, but your perseverance makes all the difference 🎉💫. Keep up the good work!
To prove this equality, you can start by expressing cot as tan:
Which gives you this:
$$\frac{1-cotθ}{1+cotθ}$$
$$\frac{1-\frac{1}{tanθ}}{1+\frac{1}{tanθ}}$$
Then, you can transform the numbers 1 so that they have the same denominator as the fraction:
$$1 = \frac{tanθ}{tanθ}$$
So, we can replace the 1s:
$$\frac{\frac{tanθ}{tanθ}-\frac{1}{tanθ}}{\frac{tanθ}{tanθ}+\frac{1}{tanθ}}$$
And add/subtract the fractions that are now on the same denominator :
$$\frac{\frac{tanθ-1}{tanθ}}{\frac{tanθ+1}{tanθ}}$$
We rewrite this big fraction:
$$\frac{tanθ-1}{tanθ} \div \frac{tanθ+1}{tanθ} $$
And we transform the division by multiplying the inverse:
$$\frac{tanθ-1}{tanθ} \times \frac{tanθ}{tanθ+1} $$
Finally, we perform the multiplication! To do this, we simplify the terms that are found in the numerator and denominator.
$$\frac{(tanθ-1)(tanθ)}{(tanθ)(tanθ+1)}$$
$$\frac{tanθ-1}{tanθ+1}$$
There you go! :D I hope this is clearer for you. If you have any other questions, don't hesitate to write us again! :)
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