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Can you write this expression as : sin@ + sqrt(sec^4 a) = sec^4 a ? (@ = theta a = alpha)
Deja vu.

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Other answers:

the key is in the infinite.
|dw:1359202205624:dw|
\[\sec^4 \alpha = \sin\theta + \sec^3 \alpha\]Nice
So, sin(theta) =sec^4(alpha) -sec^3(alpha)
Nice!
|dw:1359204540645:dw|
how this came
put y = sec ^4 theta
|dw:1359205220430:dw|
how, i didn;t understand
\[\sin \theta = \sec^4\theta \pm \sec^2 \theta \]
its secθ or secalpha
I guess I did a mistake
Sorry Alpha
|dw:1359205684540:dw|
|dw:1359205682988:dw|
|dw:1359205748804:dw|
x^2-x+x=sec^4alpha x^2=sec^4alpha x=sec^2alpha
if we put the value of sintheta in given equation, we get, |dw:1359205875734:dw|
x^2 =sec^4alpha x=+- sec^2 alpha
yes, u r right after that putting the value of x in x^2-x=sintheta
yep
answer is sec^2alpha tan^2alpha

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