Prove the following identity
cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)]
cosec (theta) + tan (theta)
= 1 / sin (theta) + (sin (theta)/cos (theta))
= [cos (theta) + sin ^2 (theta) ] / sin (theta) cos (theta)
I require help for further steps
If I substitute sin^2 (theta) = 1 - cos^2 (theta)
then:
= [cos (theta) + 1 - cos^2 (theta) ] / sin (theta) cos (theta)
With the above expression I am unable to prove the identity.
This is where I'm stuck.
Check your question. You cannot prove it because it is false.
$\displaystyle \csc(\theta)+\tan(\theta)\neq \frac{1}{\csc(\theta)-\tan(\theta)}$
If you don't believe me ,then look:
$\displaystyle \csc(\theta)+\tan(\theta)=\frac{1}{\csc(\theta)-\tan(\theta)}$
$\displaystyle (\csc(\theta)+\tan(\theta))(\csc(\theta)-\tan(\theta))=1$
$\displaystyle \csc^2(\theta)-\tan^2(\theta)=1$
$\displaystyle \csc^2(\theta)=1+\tan^2(\theta)$
$\displaystyle \csc^2(\theta)=\sec^2(\theta)$ (which is wrong!!)
According to me your question should read : Prove $\displaystyle \csc(\theta)+\cot(\theta)=\frac{1}{\csc(\theta)-\cot(\theta)}$
Hello, Ilsa!
The statement is not true. .Is there a typo?
$\displaystyle \text{Prove: }\: \csc\theta + \cot\theta \:=\:\frac{1}{\csc\theta - \cot\theta}$ . This is true
Multiply the right side by $\displaystyle \frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}$
. . $\displaystyle \frac{1}{\csc\theta - \cot\theta} \cdot\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta} \;\;=\;\;\frac{\csc\theta + \cot\theta}{\underbrace{\csc^2\!\theta - \cot^2\!\theta}_{\text{This is 1}}} \;\;=\;\; \csc\theta + \cot\theta $