Trigonometry

Ygmmumbai said:
Hi need help pls!!!

1+2sin^theta /1+3sin^theta=cos^theta
Click to expand...

Does your expression look like:

\(\displaystyle \displaystyle{\frac{1 + 2sin(\theta)}{1 + 3sin(\theta)} \ = \ cos(\theta)}\)
 
Hi thanks for your revert!

have used ^ to indicate square

so my question is 1+ 2 sine square theta divide by 1+ 3 tan square theta equals to cos square theta .

im sorry if I was not clear but it's my first query !

thanks in advance for your help !
 
"^2" for a square is clearer!

So it is \(\displaystyle \frac{1+ 2 sin^2(\theta)}{1+ 3tan^2(\theta)}= cos(\theta)\)?
That "tan" is right? You had two "sin"s before.
 
Subhotosh Khan said:
Does your expression look like:

\(\displaystyle \displaystyle{\frac{1 + 2sin^2(\theta)}{1 + 3tan^2(\theta)} \ = \ cos^2(\theta)}\)
Click to expand...

You should have then written it as:

[1 + 2sin^2(theta)]/[1 + 3tan^2(theta)] = cos^2(theta)

Those [] are very important to indicate order of operations.

What do you need to do?

Prove the above equation as identity?

or

Solve for the value of "theta" for which the equation holds true?

What are your thoughts?

If you are stuck at the beginning tell us and we'll start with the definitions.

Please share your work with us ... even if you know it is wrong

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