How can I quickly get $\sin\beta\sec\beta\cot\beta$ from $(1-\sin^2\beta)(1+\tan^2\beta)$?

Question

How I can get: $$\sin\beta\sec\beta\cot\beta$$ from: $$(1-\sin^2\beta)(1+\tan^2\beta)$$

Well, my teacher has taught me that I should only allow algebraic changes on only one side of the equation, in this case:

$$(1- \sin^2\beta)(1+\tan^2\beta) = \sin\beta\sec\beta\cot\beta$$

I have to work with the left side of the equation,

I tried:

$$(1-\sin^2\beta) = \cos^2\beta$$

$$(1+\tan^2\beta) = \sec^2\beta$$

Therefore,

$$\cos^2\beta\sec^2\beta =?$$

But from here, I have not managed to get anything closer.

So, how can I get to the desired algebraic form? What is the fastest way to get to the demonstration in these cases?

Answer 1

Recall that $$\begin{align} \cos\beta &= \dfrac{\sin\beta}{\tan\beta} \\ \sec\beta &= \dfrac1{\cos\beta} \\ \cot\beta &= \dfrac1{\tan\beta} \end{align}$$ Then, your expression can be re-written as $$\begin{align} \cos^2\beta\sec^2\beta &= \cos\beta\cos\beta\sec\beta\dfrac1{\cos\beta} \\ &= \require{cancel}\cancel{\cos\beta}\cos\beta\sec\beta\dfrac{1}{\cancel{\cos\beta}} \\ &= \cos\beta\sec\beta \\ &= \dfrac{\sin\beta}{\tan\beta}\sec\beta \\ &= \sin\beta\sec\beta\cot\beta \end{align}$$

Answer 2

We want to show

$$\sin{\beta}\sec{\beta}\cot{\beta} = (1-\sin^2{\beta})(1+\tan^2{\beta})$$

Recall $\sec^2{\beta}=\frac{1}{\cos^2{\beta}}$, so (from your working):

$$\text{RHS } = cos^2{\beta}\sec^2{\beta}=\frac{\cos^2{\beta}}{\cos^2{\beta}}=1.$$

Now consider the LHS:

$$\sin{\beta}\sec{\beta}\cot{\beta} = \sin{\beta}\frac{1}{\cos{\beta}}\frac{\cos{\beta}}{\sin{\beta}} = 1 = \text{ RHS}$$

and we are done.

Answer 3

For these type of problems, I write everything in terms of sine and cosine. Usually, the only other identity needed is $\sin^2(x)+\cos^2(x) = 1$.

For the first:

$sin(b)sec(b)Cot(b) =sin(b)\dfrac1{\cos(b)}\dfrac{\cos(b)}{\sin(b)} =\dfrac{\sin(b)\cos(b)}{\cos(b)\sin(b)} =1$.

For the second:

$(1- \sin^2(b))(1+\tan^2(b)) =\cos^2(b)(1+\dfrac{\sin^2(b)}{\cos^2(b)}) =\cos^2(b)\dfrac{\cos^2(b)+\sin^2(b)}{\cos^2(b)} =\cos^2(b)\dfrac{1}{\cos^2(b)} =1 $.

Since both are equal to $1$, they are equal.

Now that we've got this, let's run one of them in reverse.

$\begin{array}\\ (1- \sin^2(b))(1+\tan^2(b)) &=\cos^2(b)(1+\dfrac{\sin^2(b)}{\cos^2(b)})\\ &=\cos^2(b)\dfrac{\cos^2(b)+\sin^2(b)}{\cos^2(b)}\\ &=\cos^2(b)\dfrac{1}{\cos^2(b)}\\ &=1\\ &=\dfrac{\sin(b)\cos(b)}{\cos(b)\sin(b)}\\ &=\sin(b)\dfrac1{\cos(b)}\dfrac{\cos(b)}{\sin(b)}\\ &=\sin(b)\sec(b)\cot(b)\\ \end{array} $