Solve Math Problem: Simplify \sin\theta\sec\theta+\cos\theta\csc\theta

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Homework Statement


simplify:
[tex]\sin\theta\sec\theta+\cos\theta\csc\theta[/tex]


Homework Equations


Reciprocal identities, Quotient identities, Pythagorean identities


The Attempt at a Solution



[tex]\sin\theta\sec\theta+\frac{1}{\sec\theta}\frac{1}{\sin\theta}[/tex]

[tex]\sin\theta\sec\theta+\frac{1}{\sin\theta\sec\theta}[/tex]

and this is where i get stuck...can i get help from anyone?
 
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so i get...

[tex]\sin\theta\frac{1}{\cos\theta}+cos\theta\frac{1}{\sin\theta}[/tex]

then i put each in one fraction right?

[tex]\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}[/tex]

to

[tex]\tan\theta+cot\theta[/tex]

is that the simplest it can get?
 


im sorry I am really bad with these i just started doing them um so i do

[tex]\frac{\sin^2\theta+\cos\theta}{\cos\theta\sin\theta}[/tex]

i still don't get it won't that make it more complicated? :confused:
 


would it be...

[tex]\frac{\sin^2\theta+\cos^2\theta}{\cos\theta\sin\theta}[/tex]

then

[tex]\frac{1-\cos^2+\cos^2\theta}{\cos\theta\sin\theta}[/tex]

into

[tex]\frac{1}{\cos\theta\sin\theta}[/tex]

i feel so frustrated :confused: sry
 


Looks like that's the simplest you can get it.
 


Bohrok said:
Looks like that's the simplest you can get it.

[tex]2cosec(2\theta)[/tex] would seems better.
 


It depends if the OP has been exposed to double-angles yet. Trigonometric simplifications of this form are commonly taught before the student ever learns that [itex]2sin\theta cos\theta=sin(2\theta)[/itex]