Question 602392
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\cot B}{\sec B\ -\ \tan B}\ -\ \frac{\cos B}{\sec B\ +\ \tan B}\ \equiv^?\ \sin B\ +\ \csc B]


The LCD in the LHS is the product of the two denominators.  The two denominators are a conjugate pair, hence the product is the difference of two squares.  Apply the LCD:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(\sec B\ +\ \tan B)\cot B\ -\ (\sec B\ -\ \tan B)\cos B}{\sec^2 B\ -\ \tan^2 B}\ \equiv^?\ \sin B\ +\ \csc B]


Use the fact that tangent is sine over cosine, cotangent is the reciprocal of tangent, secant is the reciprocal of cosine, and cosecant is the reciprocal of sine to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(\frac{1}{\cos B}\ +\ \frac{\sin B}{\cos B})\frac{\cos B}{\sin B}\ -\ (\frac{1}{\cos B}\ -\ \frac{\sin B}{\cos B})\cos B}{\frac{1}{\cos^2 B}\ -\ \frac{\sin^2 B}{\cos^2 B}}\ \equiv^?\ \sin B\ +\ \csc B]


Simplify the LHS numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(\frac{1}{\sin B}\ +\ 1)\ -\ (1\ -\ \sin B)}{\frac{1}{\cos^2 B}\ -\ \frac{\sin^2 B}{\cos^2 B}}\ \equiv^?\ \sin B\ +\ \csc B]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{1}{\sin B}\ +\ \sin B}{\frac{1}{\cos^2 B}\ -\ \frac{\sin^2 B}{\cos^2 B}}\ \equiv^?\ \sin B\ +\ \csc B]


Combine the fractions in the LHS denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{1}{\sin B}\ +\ \sin B}{\frac{1\ -\ \sin^2 B}{\cos^2 B}}\ \equiv^?\ \sin B\ +\ \csc B]


Use the Pythagorean identity in the LHS denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\sin B}\ +\ \sin B\ \equiv^?\ \sin B\ +\ \csc B]


Finally, apply cosecant is the reciprocal of sine


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc B\ +\ \sin B\ \equiv \ \sin B\ +\ \csc B]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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