# Trigonometric Identities

• Oct 23rd 2011, 10:26 AM
Ilsa
Trigonometric Identities
Prove the following identity
cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)]
• Oct 23rd 2011, 11:05 AM
SammyS
Re: Trigonometric Identities
Quote:

Originally Posted by Ilsa
Prove the following identity
cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)]

What have you tried?

Where are you stuck?

Work on the left side. Write cosecant and tangent functions in terms of sine and cosine. Combine into one rational expression.
• Oct 23rd 2011, 11:40 PM
Ilsa
Re: 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, 01:06 AM
sbhatnagar
Re: Trigonometric Identities
Quote:

Originally Posted by Ilsa
Prove the following identity
cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)]

Check your question. You cannot prove it because it is false.

$\csc(\theta)+\tan(\theta)\neq \frac{1}{\csc(\theta)-\tan(\theta)}$

If you don't believe me ,then look:

$\csc(\theta)+\tan(\theta)=\frac{1}{\csc(\theta)-\tan(\theta)}$

$(\csc(\theta)+\tan(\theta))(\csc(\theta)-\tan(\theta))=1$

$\csc^2(\theta)-\tan^2(\theta)=1$

$\csc^2(\theta)=1+\tan^2(\theta)$

$\csc^2(\theta)=\sec^2(\theta)$ (which is wrong!!)(Headbang)

According to me your question should read : Prove $\csc(\theta)+\cot(\theta)=\frac{1}{\csc(\theta)-\cot(\theta)}$
• Nov 4th 2011, 07:06 AM
Soroban
Re: Trigonometric Identities
Hello, Ilsa!

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

Quote:

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

Multiply the right side by $\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}$

. . $\frac{1}{\csc\theta - \cot\theta} \cdot\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta} \;\;=\;\;\frac{\csc\theta + \cot\theta}{\underbrace{\csc^2\!\theta - \cot^2\!\theta}_{\text{This is 1}}} \;\;=\;\; \csc\theta + \cot\theta$