[গণিত] হলে, উদ্দীপক হতে দেখাও যে, [গণিত] 2 [গণিত] - [গণিত] 2 [গণিত] =4.

দেওয়া আছে, C = 75° ∴ A = 1800 –750 – 650 = 400 [গণিত]

[গণিত] হলে, উদ্দীপক হতে দেখাও যে, [গণিত] 2 [গণিত] - [গণিত] 2 [গণিত] =4.

দেওয়া আছে, C = 75° ∴ A = 1800 –750 – 650 = 400 [গণিত]

প্রশ্ন পাওয়া যায়নি

HSC · Higher Math 1st Paper · Chapter 7: সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত · সৃজনশীল

হলে, উদ্দীপক হতে দেখাও যে, 2 - 2 =4.

উদ্দীপক

Rajshahi · 2022

উত্তর

\begin{aligned}& \cos \alpha \cos \beta-\sin \alpha \sin \beta=1 \\& \Rightarrow \cos (\alpha+\beta)=1 \\& \Rightarrow \alpha+\beta=0^0\end{aligned}

উত্তর

\begin{aligned} & \text{ L.S }=\sec 2 A-\sqrt{3} \operatorname{\cosec} 2 A \\ & =\frac{1}{\cos 2 A}-\frac{\sqrt{3}}{\sin 2 A} \\ & =\frac{\sin 2 A-\sqrt{3} \cos 2 A}{\sin 2 A \cos 2 A} \\ & =2 \times \frac{\frac{1}{2} \sin 2 A-\frac{\sqrt{3}}{2} \cos 2 A}{\sin 2 A \cos 2 A} \\ & =2 \times 2 \frac{\cos \frac{\pi}{3} \sin 2 A-\sin \pi / 3 \cos 2 A}{2 \sin 2 A \sin \cos 2 A} \\ & =4 \frac{\sin (2 A-\pi / 3)}{\sin (2 \times 2 A)}[\because \sin A \cos B-\cos A \sin B=\sin A-B] \\ & =4 \frac{\sin \left(80^{\circ}-60^{\circ}\right)}{\sin 160^{\circ}} \\ & =4 \frac{\sin 20^{\circ}}{\sin \left(180^{\circ}-20^{\circ}\right)} \\ & =4 \frac{\sin 20^{\circ}}{\sin 20^{\circ}} \\ & =4(\operatorname{showed~})\end{aligned}

উত্তর

\begin{aligned} \text{ L. S. }= & \frac{a^2}{b c} \sin (B-c)+\frac{b^2}{c a} \sin (c-A)+\frac{c^2}{a b} \sin (A-B) \\ = & \frac{a^2}{b c}(\sin B \cos C-\cos B \sin C)+\frac{b^2}{c a}(\sin C \cos A-\cos C \sin A)+\frac{c^2}{a b}(\sin A \cos B \ldots \cos A \sin B) \\ & \therefore \sin (A-B)=\sin A \cos B-\sin B) \\ = & \frac{a^2}{b c}\left(\frac{b}{2 R} \cos C-\frac{c}{2 R} \cos B\right)+\frac{b^2}{c a}\left(\frac{c}{2 R} \cos A-\frac{a}{2 R} \cos A C\right)+\frac{c^2}{a b}\left(\frac{a}{2 R} \cos B-\frac{b}{2 R} \cos A\right) \\ = & \frac{a^2}{b C 2 R}(b \cos C-c \cos B)+\frac{b^2}{c a 2 R}(\cos A-a \cos C)+\frac{c^2}{a b 2 R}(a \cos B-b \cos A) \\ & \quad \frac{a}{2 R b C}(b \cos C-c \cos B)(b \cos C+c \cos B)+\frac{b}{2 R c a}(c \cos A-a \cos C)(\cos A+a \cos C) \\ & \quad+\frac{c}{2 R a B}(a \cos B-b \cos A)(a \cos B+b \cdot \cos A) \\& =\frac{a}{2 R b c}\left(b^2 \cos^2 c-c^2 \cos^2 B\right)+\frac{b}{2 R c a}\left(c^2 \cos^2 A-a^2 \cos^2 C\right)+\frac{c}{2 k a b}\left(a^2 \cos^2 B-b^2 \cos^2 A\right) \\ & =\frac{a \times a}{2 R \times 4 R \Delta}\left(b^2 \cos^2 c-c^2 \cos^2 B\right)+\frac{b \times b}{2 R \times \Delta 4 R}\left(c^2 \cos^2 A-a^2 \cos^2 C\right)+\frac{c \times c}{2 R \times \Delta 4 R}\left(a^2 \cos^2 B-\right. \\ & \left.b^2 \cos^2 A\right) \because \frac{a b c}{4 R}=A \\ & =\frac{1}{8 R^2 \Delta}\left(a^2 b^2 \cos^2 c-a^2 c^2 \cos^2 B+b^2 c^2 \cos^2 A-a^2 b^2 \cos^2 C+a^2 c^2 \cos^2 B-b^2 c^2 \cos^2 A\right) \\ & =\frac{1}{8 R^2 D} \times 0=0 \\ & =R \cdot S \text{ (Proved) }\end{aligned}

HSCHigher Math 1st PaperChapter 7: সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাতRajshahi

12টি সম্পর্কিত প্রশ্ন — উত্তর ও ব্যাখ্যা লকড

1. $\sqrt{3}$

ক $\pi$
খ $\pi$
গ $\pi$
ঘ $\pi$