Solve the equation 1 ( cos &#x2061;<!-- ⁡ --> &#x03B8;<!

If ${\displaystyle \sin x+\sin y=a\text{and}\cos x+\cos y=b}$ then find ${\displaystyle \tan (x-\frac{y}{2})}$

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 11 passengers per minute.

Proving ${\displaystyle {({\sin }^2\alpha +\sin \alpha \cos \alpha )}^{\sin \alpha }{({\cos }^2\alpha +\sin \alpha \cos \alpha )}^{\cos \alpha }\le 1}$

Prove the following identity: $\frac{1}{(\sec A+\tan A)}=\sec A-\tan A$

Why is ${\displaystyle \tan \theta \approx \frac{1}{\frac{\pi }{2}-\theta }}$ for ${\displaystyle \theta }$ close to ${\displaystyle \frac{\pi }{2}}$?
I wanted to see what the behaviour of the steep part of the ${\displaystyle \tan }$ curve was like, i.e. the behaviour of ${\displaystyle \tan \left(x\right)asx\to {\left(\frac{\pi }{2}\right)}^{-}}$. So by thinking about a shift of the graph of ${\displaystyle \tan \left(x\right)by\frac{\pi }{2}}$ to the left, I put some small (positive and negative) values of ${\displaystyle \theta }$ into my calculator for the function ${\displaystyle \tan (\theta +\frac{\pi }{2})}$
${\displaystyle \tan \theta \approx \frac{1}{\frac{\pi }{2}-\theta }}$ for ${\displaystyle \theta }$ close to ${\displaystyle \frac{\pi }{2}}$?
or, in more colloquial terms,
The steep part of ${\displaystyle \tan x}$ is just like the steep part of ${\displaystyle \frac{1}{x}}$
But why is this the case? I couldn't deduce it easily using the Maclaurin expansion of ${\displaystyle \tan \left(x\right)}$. Is there a more intuitive explanation? I couldn't think of any explanations analogous to those explaining small angle approximations.

Solve ${\displaystyle {\cot }^2\theta -{\cos }^2\theta }$

Use a half-angel formula to find the exact value of ${\displaystyle \tan 15}$