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Question Number 176211 by adhigenz last updated on 15/Sep/22

	Answered by mahdipoor last updated on 15/Sep/22
	$2021((cos(a+b))/(cosa))sinb=((sina)/(cosa)) 2021((cosa.cosb−sina.sinb)/(cosa))sinb=tana 2021cosb.sinb−2021tana.sin^2 b=tana ⇒tana=((2021cosb.sinb)/(1+2021sin^2 b))=f(b) max f = max tana { ((cosb.sinb=((sin2b)/2))),((sin^2 b=((1−cos2b)/2))) :}⇒ f(b)=((sin(2b))/(((2023)/(2021))−cos(2b))) (get((2023)/(2021))=c) (df/db)=0 ⇒ cos(2b)=(1/c) ⇒ sin(2b)=((√(c^2 −1))/c) max f=(((√(c^2 −1))/c)/(c−1/c))=((√(c^2 −1))/(c^2 −1))=((2021(√(2022)))/(4044))$
	$2021 \frac{c o s (a + b)}{c o s a} s i n b = \frac{s i n a}{c o s a}$ $2021 \frac{c o s a . c o s b - s i n a . s i n b}{c o s a} s i n b = t a n a$ $2021 c o s b . s i n b - 2021 t a n a . {s i n}^{2} b = t a n a$ $\Rightarrow t a n a = \frac{2021 c o s b . s i n b}{1 + 2021 {s i n}^{2} b} = f (b)$ $m a x f = m a x t a n a$ ${\begin{cases} c o s b . s i n b = \frac{s i n 2 b}{2} \\ {s i n}^{2} b = \frac{1 - c o s 2 b}{2} \end{cases} \Rightarrow$ $f (b) = \frac{s i n (2 b)}{\frac{2023}{2021} - c o s (2 b)} (g e t \frac{2023}{2021} = c)$ $\frac{d f}{d b} = 0 \Rightarrow c o s (2 b) = \frac{1}{c} \Rightarrow s i n (2 b) = \frac{\sqrt{c^{2} - 1}}{c}$ $m a x f = \frac{\sqrt{c^{2} - 1} / c}{c - 1 / c} = \frac{\sqrt{c^{2} - 1}}{c^{2} - 1} = \frac{2021 \sqrt{2022}}{4044}$
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