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50 points Simplify the expression. sin^2 theta/csctheta tan^2theta a. sin^2 theta b.sin cos^2 theta c.sec^2 theta d. csc theta

User Ilijanovic
by
4.4k points

2 Answers

2 votes

Answer:

B. sin(theta) cos^2(theta)

Explanation:

sin^2(theta) / csctheta tan^2(theta)

To solve, we convert to sine and cosine and then simplify.

means by the priority of operators (* and / from left to right)

(sin^2(theta) / csctheta) * tan^2(theta)

= sin^2(theta) / (1/sin(theta) * (sin(theta)/cos(theta)^2

= sin^2(theta) * sin(theta) * sin^2(theta) / cos^2(theta)

= sin^5(theta) / cos^2(theta)

which obviously does not correspond to any of the answers.

IF the expression were missing parentheses in the denominator, then it becomes:

sin^2(theta) / (csctheta tan^2(theta))

= sin^2(theta) / (1/sin(theta) * sin^2(theta) /cos^2(theta))

= sin^2(theta) / (sin(theta)/cos^2(theta)

= sin(theta) cos^2(theta)

which corresponds to answer B.

User Sahith Kurapati
by
4.3k points
4 votes

Answer:


\Large \boxed{\bold{B.} \ \mathrm{sin(\theta) \cdot cos^2(\theta) }}

Explanation:


\displaystyle \mathrm{(sin^2(\theta ) )/( csc(\theta) \cdot tan^2(\theta)) }

Simplify the expression.


\displaystyle \mathrm{(sin^2(\theta ) )/((1)/(sin) (\theta) \cdot ( sin^2(\theta))/( cos^2(\theta)) ) }


\displaystyle \mathrm{(sin^2(\theta ) )/((sin^2 )/(sin) (\theta) \cdot cos^2(\theta) ) }


\displaystyle \mathrm{(sin^2(\theta ) )/(sin (\theta) \cdot cos^2(\theta) ) }


\displaystyle \mathrm{(sin^2(\theta ) )/(sin (\theta) ) \cdot cos^2(\theta) }


\displaystyle \mathrm{sin(\theta) \cdot cos^2(\theta) }

User Victor Rendina
by
4.7k points