Having a mental block and can't seem to work out these questions (tech free) 
The first two I can simplify partially but not fully to get the stated answers, the last two I'm just lost so any insight on how to solve them would be much appreciated.
- cos^2(x)/sin(x) + sin(x) answer=cosec(x)
First one:
}{sin(x)}+sin(x))
Take sin(x) out as a factor
\left(\frac{cos^2(x)}{sin^2(x)}+1\right))
(cot^2(x)+1))
+1 = csc^2(x))
\cdot csc^2(x))
)
Slightly shorter method
}{\sin(x)}+\sin(x)=\frac{\cos^{2}(x)+\sin^{2}(x)}{sin(x)}=\frac{1}{\sin(x)}=\csc(x))
(as
+\sin^{2}(x)=1)
)
For the others use
to find sin(x), then tan(x)=sin(x)/cos(x), then sec(x)=1/cos(x)
For sin(x+y), you've found cos(x) and sin(x) already, draw up a triangle with the opposite=5 and adjacent=12 with angle y so that you can find cos(y) an sin(y), then you can use the addition formula sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
Then for tan(x+2y) we know that tan(a+b)=
=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)})
, so we need tan(2y), which we can use the double angle formula for,
=\frac{2\tan(y)}{1-\tan^{2}(y)})
so then use
=\frac{\tan(x)+\tan(2y)}{1-\tan(x)\tan(2y)})
and sub in the values.
For tan(x)=2, draw up a traingle with opposite=2, adjacent=1, then find the hypotenus=root(4+1)=root(5). Now you can find sin(x)=2/root(5) and cos(x)=1/root(5). Now you can use the double angle formulas
=2\sin(x)\cos(x))
and
=1-2\sin^{2}(x))
and take into account the quadrant.
For the last one
}{1-\tan^{2}(\frac{x}{2})})
, is just like you double angle formula for tan, so we can say that
}{1-\tan^{2}(\frac{x}{2})}=\tan(x))
Hope that helps
(and hope there aren't any mistakes there, eyes are tired so there might be)
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