For all values of θ, 1 + cos θ is always_
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Answer:
For all values of [tex]\sf \: \theta [/tex], 1 + [tex]\sf \: cos\theta [/tex] is always greater than or equals to 0 but smaller than or equals to 2.
Step-by-step explanation:
Given that,
[tex]\sf \: \theta \: \in \: R \\ [/tex]
We know,
[tex]\sf \: cos \theta \: \in \: [ - 1, \: 1]\\ [/tex]
[tex]\sf \: - 1 \leqslant \: cos \theta \: \leqslant 1\\ [/tex]
On adding 1 in each term, we get
[tex]\sf \: - 1 + 1 \leqslant \: 1 + cos \theta \: \leqslant 1 + 1\\ [/tex]
[tex]\sf \: 0 \leqslant \: 1 + cos \theta \: \leqslant 2\\ [/tex]
[tex]\implies\sf \: 1 + cos \theta \: \in \: [ 0, \: 2] \\ [/tex]
Hence,
For all values of [tex]\sf \: \theta [/tex], 1 + [tex]\sf \: cos\theta [/tex] always is greater than or equals to 0 but smaller than or equals to 2.
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \: \in \: ( - \infty , \infty )\\ \\ \sf y = cosec & \sf y \leqslant - 1 \: \: or \: \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant - 1 \: \: or \: \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \: \in \: ( - \infty , \infty ) \end{array}} \\ \end{gathered} \\ [/tex]