Question

How do you simplify this?

Answer 1

\[\frac{ cscx(\sin^2x + \cos^2xtanx) }{ sinx + cosx }\]

Answer 2

this is not as hard as it looks, and has nothing really to do with trig
it is algebra
i can show you if you like, but it will be easy if you replace \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\) and then consider the expression
\[\frac{\frac{a}{b}\left(b^2+a^2\frac{b}{a}\right)}{b+a}\]

Answer 3

clear or no?

Answer 4

somewhat, how would you distribute the a/b ?

Answer 5

yes
then there will be a raft of cancellation

Answer 6

oh how?

Answer 7

\[\frac{\frac{a}{b}\left(b^2+a^2\frac{b}{a}\right)}{b+a}\]
\[=\frac{\frac{a}{b}\times b^2+ a^2\frac{b}{a}\times \frac{a}{b}}{b+a}\]

Answer 8

then cancel in the top and get
\[\frac{ab+a^2}{b+a}\]

Answer 9

then factor the numerator and you can cancel again

Answer 10

clear on those steps? if not let me know

Answer 11

yes, I got cos

Answer 12

got it

Answer 13

yes, can i substitute a and b in \[\frac{ 6\sin \phi}{ \cos^2 \phi }\ \times \frac{ \cos^2 \phi - \cos \phi \sin \phi }{ \cos^2 \phi - \sin^2 \phi }\]

Answer 14

that is what i would do
it makes the algebra much clearer
at some point you might have to use a trig identity, but that is usually at the end

Answer 15

i would look to factor and cancel first
don't multiply out until you do that

Answer 16

clear?

Answer 17

This is what i got
\[\frac{ 6\sin \phi }{ \cos^2 \phi }\times \frac{ 1 }{ \sin \phi }\]

Answer 18

i can try it and see if i get the same thing
but you can cancel the sine if your answer is right

Answer 19

that is not what i get
i have a \(\sin(x)+\cos(x)\) still in the denominator

Answer 20

thats what i had, you cant cancel the cos in the numerator with the one in the denominator?

Answer 21

\[\frac{6b}{a^2}\times \frac{a^2-ab}{a^2-b^2}\]
\[\frac{6b}{a^2}\times \frac{a(a-b)}{(a+b)(a-b)}\]
\[\frac{6b}{a(a+b)}\]

Answer 22

okay thats alot more clear now. what about using the sum and difference to find tan 75