Sinθ+cotθ×cosθ=cscθ

Mathematics

1 answer:

Assoli18 [71]3 years ago

5 0

Answer:

Step-by-step explanation:

$\sin \theta+\cot \theta\cos \theta=\sin \theta+\displaystyle \frac{\cos \theta}{\sin \theta}\cos \theta=\frac{\sin^2 \theta+\cos ^2 \theta}{\sin \theta}=\\\frac{1}{\sin \theta}=\csc \theta$

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What will be the value of

madreJ [45]

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

$\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}$

Then you want to know the value of

$\displaystyle\lim_{n\to\infty}a_n$

if it exists.

To show the limit exists and that $a_n$ converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that $a_1=\sqrt{42}\le\sqrt{49}=7$ . Suppose $a_k\le 7$ . Then $a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7$ . So by induction, $a_n$ is bounded above by 7 for all $n$ .

Monontonicity: We have $a_1=\sqrt{42}$ and $a_2=\sqrt{42+\sqrt{42}}$ . It should be quite clear that $a_2>a_1$ . Suppose $a_k>a_{k-1}$ . Then $a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k$ . So by induction, $a_n$ is monotonically increasing.