Mathematics, 02.08.2019 08:30 edith47

Verify that seca + tana = (cos)/(1-sina)

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Refresh if you see display errors such as \begin{aligned}, etc.====== $\sec a + \tan a = \dfrac{\cos a}{1-\sin a}$ focusing on the left-hand side $\text{lhs} = \sec a + \tan a$ one strategy is to rewrite everything in terms of sine and cosine.since $\sec a = \frac{1}{\cos a}$ and $\tan a = \frac{\sin a}{\cos a}$ , we have $\begin{aligned} \text{lhs} & = \sec a + \tan a \\ & = \frac{1}{\cos a} + \frac{\sin a}{\cos a} \\ & = \frac{1 + \sin a}{\cos a} \end{aligned}$ they have the same denominator, which is why i am able to combine them.that looks as simple as simple gets. can leave it alone. (read the remark at the end, because there are things you can do here, too.)======focusing on the right-hand side $\text{rhs} = \dfrac{\cos a}{1-\sin a}$ in this situation, there is one strategy we can use: multiplication by the conjugate.for the right hand side, the conjugate of $1-\sin a$ is $1+\sin a$ . we can multiply the numerator and denominator of the right hand side by this conjugate.why do this? because it will produce a difference of squares in the denominator, which may permit us to use a pythagorean identity.(another rationale is that, if you look on the left side, we have 1+sina; this can lead to you conclude that we need to multiply by 1+sina on the right side to get it there as well.) $\begin{aligned} \text{rhs} & = \dfrac{\cos a}{1-\sin a} \cdot \frac{1+\sin a}{1+\sin a} \\ & = \dfrac{\cos a(1+\sin a)}{(1-\sin a)(1+\sin a)} \end{aligned}$ the denominator is a difference of squares expansion, in the form of $(a-b)(a+b) = a^2 - b^2$ . doing this to the denominator gets us $\begin{aligned} \text{rhs} & = \dfrac{\cos a(1+\sin a)}{(1-\sin a)(1+\sin a)} \\ & = \dfrac{\cos a(1+\sin a)}{1-\sin^2 a} \end{aligned}$ the denominator is a variation of the pyth. identity.with $\sin^2 a + \cos^2 a = 1$ , subtracting both sides by $\sin^2 a$ gets us $\cos^2 a= 1 -\sin^2 a$ . hence $\begin{aligned} \text{rhs} & = \dfrac{\cos a(1+\sin a)}{1-\sin^2 a} \\ & = \dfrac{\cos a(1+\sin a)}{\cos^2 a} \end{aligned}$ the cosines (partially) cancel out and we are left with $\begin{aligned} \text{rhs} & = \dfrac{1+\sin a}{\cos a} \\ & = \text{lhs} \end{aligned}$ ======you could have also did the conjugate step for the left-hand side when you finished simplifying it all. there are usually many ways to prove an identity.i hope i'm not doing your entire homework packet or something like that (though i do understand that this one involves something different than the other questions you've posted)

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Answer from: Quest

No a triangle cannot be formed because when you add the two smallest number together (which would be 11) the have to be bigger than the number it is "against" ( which would be 12)

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