Prove the following identity

cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)]

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- Oct 23rd 2011, 09:26 AMIlsaTrigonometric Identities
Prove the following identity

cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)] - Oct 23rd 2011, 10:05 AMSammySRe: Trigonometric Identities
- Oct 23rd 2011, 10:40 PMIlsaRe: Trigonometric Identities
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. - Oct 24th 2011, 12:06 AMsbhatnagarRe: Trigonometric Identities
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!!)(Headbang)

According to me your question should read : Prove $\displaystyle \csc(\theta)+\cot(\theta)=\frac{1}{\csc(\theta)-\cot(\theta)}$ - Nov 4th 2011, 06:06 AMSorobanRe: Trigonometric Identities
Hello, Ilsa!

The statement is not true. .Is there a typo?

Quote:

$\displaystyle \text{Prove: }\: \csc\theta + \cot\theta \:=\:\frac{1}{\csc\theta - \cot\theta}$ . 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 $