# Thread: Identities

1. ## Identities

Can somebody help me verify these two identities
$\displaystyle (csc t - cot t)^4 (csc t + cot t)^4=1$
$\displaystyle (csc t + cot t)^2(csc t - cot t)^2(csc t + cot t)^2 (csc t -cot t)^2$
$\displaystyle (csc t +cot t)^2(1)^2(csc t + cot t)^2(1)^2$
Is this one right so far, what do I do next?

$\displaystyle \frac{tan a}{1+sec a}+\frac{1+sec a}{tan a}= 2 csc a$
How do i do this one?

2. Originally Posted by purplec16
Can somebody help me verify these two identities
$\displaystyle (csc t - cot t)^4 (csc t + cot t)^4=1$

$\displaystyle \frac{tan a}{1+sec a}+\frac{1+sec a}{tan a}= 2 csc a$
How do i do this one?
$\displaystyle (\csc{t} - \cot{t})^4 (\csc{t} + \cot{t})^4=$

$\displaystyle [(\csc{t} - \cot{t})(\csc{t} + \cot{t})]^4=$

$\displaystyle [\csc^2{t} - \cot^2{t}]^4 =$

$\displaystyle [(1+\cot^2{t}) - \cot^2{t}]^4 =$

$\displaystyle 1^4 = 1$

$\displaystyle \frac{\tan{a}}{1+\sec {a}}+\frac{1+\sec{a}}{\tan{a}}=$

multiply both fractions by $\displaystyle \frac{\cos{a}}{\cos{a}}$ ...

$\displaystyle \frac{\sin{a}}{\cos{a}+1}+\frac{\cos{a}+1}{\sin{a} }=$

common denominator and add ...

$\displaystyle \frac{\sin^2{a}}{\sin{a}(\cos{a}+1)}+\frac{\cos^2{ a} + 2\cos{a} + 1}{\sin{a}(\cos{a}+1)}=$

$\displaystyle \frac{\sin^2{a}+\cos^2{a} + 2\cos{a} + 1}{\sin{a}(\cos{a}+1)}=$

$\displaystyle \frac{2\cos{a} + 2}{\sin{a}(\cos{a}+1)}=$

$\displaystyle \frac{2(\cos{a} + 1)}{\sin{a}(\cos{a}+1)}=$

$\displaystyle \frac{2}{\sin{a}} = 2\csc{a}$

3. Thank you so much, Can you help me with two more problems
$\displaystyle sec \theta + csc \theta-cos\theta-sin\theta= sin\theta tan\theta+ cos\theta cot\theta$

and

$\displaystyle \frac{cot(-t)+tan(-t)}{cot t}= -sec^2 t$