Hi there!
And a quick digression, I’ve decided to quit business consulting at a language school on August 31, 2022. I dared use the word “business consulting” because I have been giving them scads of advice to buoy up their marketing strategies; however, they had no ears to listen to my advice.
I’ve become sick and tired of their “fuddy-duddy” operations sticking to mass marketing. More’s a pity; they are far away from innovative reform to make their betterment. Truth be told, I’m starting to smell a rat.
Generations have changed? How many people will be able to get the lay of the land with my post? Are they a bunch of dullards or people I find obnoxious in some way? I’d better keep my distance from the entity that will go the course of the dodo down the road.
Apparently, my digression ultimately has got a tangent. It is high time to find a tangent value.
Find the value of \(\tan105°\).
Here is my solution to the problem.
Given the sum identity \(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 – \tan\alpha\tan\beta}\),
\(\tan105°\)
\(= \tan(60° + 45°)\)
\(= \frac{\tan60° + \tan45°}{1 – \tan60°\tan45°}\)
\(= \frac{\sqrt{3} + 1}{1 – \sqrt{3}}\)
Then, rationalize the denominator,
\(\frac{\sqrt{3} + 1}{1 – \sqrt{3}}\)
\(= \frac{(1 + \sqrt{3})^{2}}{(1 + \sqrt{3})(1 – \sqrt{3})}\)
\(= -2 – \sqrt{3}\)
Being kind takes energy and commitment. I’ve mellowed out with ages; however, I’d better use my power for more productive things to age gracefully.
Wish that language school many more successful years.
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
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【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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