Thinking questions involve combinations of more than one strategy see how knowledge of proving identities and solving equation is tested.
Steps to solve the following equations are:
• Simplify the LHS
• Then solve the equation in the given domain

Question 1:    \(\frac{1 -cos θ}{sinθ} - \frac{sin θ}{1 + cos θ}\)  = tan θ 2,0°≤ ≤360°

Solation:  \(\frac{1 -cos θ}{sin θ} - \frac{sin θ}{1 + cos θ}\) 

=     \(\frac{\big(1 -cos θ)(1 + cos θ) - sin^2}{sin θ(1 + cos θ)}\)

=  \(\frac{1 - cos^2θ - sin^ θ}{sin θ(1 + cos θ)}\)

=    \(\frac{1 - (cos^2θ + sin^2θ}{sin θ(1 + cos θ}\)

=    \(\frac{1 - }{sin θ(1 + cos θ)}\) = 0

Since = 0 we need to solve the equation tan2=0, 0°≤ ≤360° Consider domain for 2: 0°≤2 ≤720° 2=0°,180°,360°,540°,720° =0°,90°,180°,270°,360°

Note: Period of tan2 is 90°. Continue adding multiples of 90 in the domain to get all the possible solutions to the first vale of 0°.

Question 2: Prove Identity

   \(\frac{1 +cos θ}{sin θ} + \frac{sin θ}{1 + cos θ}\) = \(\frac{2}{sin θ}\)

Solation:  Simplify the LHS

\(\frac{1 +cos θ}{sin θ} + \frac{sin θ}{1 + cos θ}\) =    \(\frac{2}{sin θ}\)

=    \(\frac{(1 + cos θ)^2θ + sin^2}{sin θ(1 + cos θ)}\)

=    \(\frac{1 +2 cos θ+ cos^2θ + sin^2}{sin ( 1 + cos θ)}\)

=    \(\frac{2 + 2 cos θ}{sin θ(1 + cos θ)}\) =    \(\frac{2}{sin θ}\)

Then Solve

\(\frac{1 +cos θ}{sin} + \frac{sin θ}{1 + cos θ}\) =    \(\frac{2}{sin θ}\) =    \(\frac{2}{cos θ}\)

Solve Equation

   \(\frac{2}{sin θ}\) =    \(\frac{2}{cos θ}\)

   \(\frac{2}{2}\)    =     \(\frac{sin θ}{cos θ}\)

1 = tan θ

=45°+180,∈

Question 3: Prove the identity

   \(\frac{cos θ}{1 - tan θ} + \frac{sin θ}{1 - cot θ}\) =    \({sin θ + cos θ}\)

Solation: \(\frac{cos θ}{1 - tan θ} + \frac{sin θ}{1 - cot θ}\) =    \({sin θ + cos θ}\)

LHS = \(\frac{cos θ}{1 - tan} + \frac{sin θ}{1 - cot θ}\)

=  \(\frac{cos θ}{1 - frac{sin θ}{cos θ}} + \frac{sin θ}{1 - \frac{cos θ}{sin θ}}\)

=  \(\frac{cos}{\frac{cos θ - sin θ}{cos θ}} + \frav{sin θ}{\frac{sin θ - cos θ}{sin θ}}\)

=    \(\frac{cos^2θ}{cos θ - sin θ} + \frac{sin^2θ}{sin θ - cos θ}\)

=    \(\frac{cos^2θ}{cos θ - sin θ} - \frac{sin^2}{cos θ - sin θ}\)

=    \(\frac{cos^2 θ - sin^2θ}{cos θ - sin θ}\) =    \(\frac{(cos θ + sin θ)(cos θ- sin θ}{cos θ - sin θ}\)

=    \({cos θ + sin θ}\) =    \({LHS}\)

Then find general solution for

\(\frac{cos θ}{1 - tan θ} + \frac{sin θ}{1 - cot θ}\) =    \({sin θ - cos θ}\)

Solution of the equation

\(\frac{cos θ}{1 - tan θ} + \frac{sin θ}{1 - cot θ}\) =    \({sin θ - cos θ}\)

   \( =    [latex{cos θ + sin θ}\)

  \( =   [latex{sin θ – cos θ}\)

is the equation to be solved

2 cos θ  = 0

cos θ = 0

θ = 90° (2+1), ∈