Simplify \$(("sin" \theta) / (1 + "cos" \theta)) + ((1 + "cos" \theta) / ("sin" \theta))\$ by using reciprocal trigonometric identity.

The given expression is

\$(("sin" \theta) / (1 + "cos" \theta)) + ((1 + "cos" \theta) / ("sin" \theta))\$

Take LCM

⇒ \$(("sin"\theta)^2 + (1 + "cos"\theta)^2)/ ("sin"\theta (1 + "cos"\theta))\$

Distribute \$1 + ("cos"\theta)^2\$ \$ ("a" + "b")^2 = "a"^2 + 2"ab" + "b"^2\$

⇒ \$("sin"^2(\theta) + (1 + 2 "cos"\theta + "cos"^2(\theta))) / ("sin"\theta (1 + "cos"\theta))\$

Use the identity \$"sin"^2 ("A") + "cos"^2("A") = 1\$.

Make it simpler the combine like terms in the numerator

⇒ \$ (1 + 1 + 2"cos"(\theta)) / ("sin"\theta (1 + "cos"\theta))\$

⇒ \$(2 + 2"cos"\theta) /("sin"\theta (1 + "cos"\theta))\$

Factor out 2 from the numerator then cancel out the common factor

⇒ \$(2(1 + "cos"(\theta)))/("sin"\theta (1 + "cos"\theta))\$

⇒ \$2/("sin"(\theta))\$

Use a reciprocal trigonometric identity

⇒ \$2/("sin"(\theta)\$

⇒ \$2"csc"(\theta)\$

Therefore the \$(("sin" \theta) / (1 + "cos" \theta)) + ((1 + "cos" \theta) / "sin" \theta)\$ = \$2"csc"(\theta)\$.

Incorrect because it’s missing the coefficient 2 in front of csc⁡θ

\$"csc" \theta\$

Wrong because it gives the negative of the correct answer, and there is no negative sign present in the simplified expression

\$-"csc" \theta\$

Incorrect: The simplified expression is 2csc⁡θ, not −2csc⁡θ

\$-2"csc"\theta\$

Correct. The calculation has been accurately completed

\$2"csc"(\theta)\$

Option

D

Can you summarize what you’ve understood in the above steps?