SOLUTION: Hi. Please help me prove that this equation is an identity because I really don't know what to do. Thank you!
(cot B / sec B - tan B) - (cos B / sec B + tan B) = sin B + csc
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Question 602392: Hi. Please help me prove that this equation is an identity because I really don't know what to do. Thank you!
(cot B / sec B - tan B) - (cos B / sec B + tan B) = sin B + csc B
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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:
\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:
\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:
\ -\ (1\ -\ \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:
Use the Pythagorean identity in the LHS denominator:
Finally, apply cosecant is the reciprocal of sine
John
My calculator said it, I believe it, that settles it
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