SOLUTION: prove that (\sin (\theta )+\tan (\theta ))/(\csc (\theta )+\cot (\theta )) = sin(\theta ) tan (\theta )

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Question 1207649: prove that (\sin (\theta )+\tan (\theta ))/(\csc (\theta )+\cot (\theta )) = sin(\theta ) tan (\theta )
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given:

manipulate left side

%28sin+%28theta+%29%2Btan+%28theta+%29%29%2F%28csc+%28theta+%29%2Bcot+%28theta+%29%29+...use identities

=

=

= ...simplify

=%28sin+%28theta+%29%2Fcos+%28theta+%29%29%2F%281%2Fsin+%28theta+%29%29

=sin+%28theta+%29%28sin+%28theta+%29%2Fcos+%28theta+%29%29

=sin+%28theta+%29%2Atan%28theta+%29=< proven


Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
By the way, that notation with backward slashes "\" is not compatible with 
this site, which is written with HTML, so there is no need using it here.

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Here's an alternate proof, using the fact that any quantity multiplied by
its reciprocal is always 1. 

Start with the left side:

%28sin%28theta%29%2Btan%28theta%29%29%2F%28csc%28theta%29%2Bcot%28theta%29%29

Multiply top and bottom by the "conjugate" of the denominator:

expr%28%28sin%28theta%29%2Btan%28theta%29%29%2F%28csc%28theta%29%2Bcot%28theta%29%29%29%22%22%2A%22%22expr%28%28csc%28theta%29-cot%28theta%29%29%2F%28csc%28theta%29-cot%28theta%29%29%29



Use the fact that multiplying an expression by its reciprocal is always 1.
Also the denominator is a well-known identity:

%28%0D%0A1-sin%28theta%29cot%28theta%29%2Btan%28theta%29csc%28theta%29-1%29%2F1

-sin%28theta%29cot%28theta%29%2Btan%28theta%29csc%28theta%29



-cos%28theta%29%2B1%2Fcos%28theta%29

%28-cos%5E2%28theta%29%2B1%29%2Fcos%28theta%29

sin%5E2%28theta%29%2Fcos%28theta%29

%28sin%28theta%29sin%28theta%29%29%2Fcos%28theta%29

sin%28theta%29%2Aexpr%28sin%28theta%29%2Fcos%28theta%29%29

sin%28theta%29%2Atan%28theta%29

Edwin