বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
f(x)=sin−1p+sin−1q+sin−1rA=cosx−cos2xR=1−cosx \begin{array}{l}f(x)=\sin ^{-1} p+\sin ^{-1} q+\sin ^{-1} r \\ A=\cos x-\cos 2 x \\ R=1-\cos x\end{array} f(x)=sin−1p+sin−1q+sin−1rA=cosx−cos2xR=1−cosx
প্রমাণ কর যে, tan−113=12sin−135\tan ^{-1} \frac{1}{3}=\frac{1}{2} \sin ^{-1} \frac{3}{5}tan−131=21sin−153
f(x)=π\mathrm{f}(\mathrm{x})=\pif(x)=π হলে দেখাও যে, p1−p2+q1−q2+r1−r2\mathrm{p} \sqrt{1-\mathrm{p}^{2}}+\mathrm{q} \sqrt{1-\mathrm{q}^{2}}+\mathrm{r} \sqrt{1-\mathrm{r}^{2}}p1−p2+q1−q2+r1−r2=2pqr
সমাধান কর : AR=1\frac{A}{R}=1RA=1 যখন 0 < x < π \pi π
উদ্দীপক-১: f(x)=cosxf(x)=\cos xf(x)=cosx
উদ্দীপক-2: cot−1(1x)+12sec−1(1+y21−y2)+12cosec−1(1+z22z)=π\cot ^{-1}\left(\frac{1}{x}\right)+\frac{1}{2} \sec ^{-1}\left(\frac{1+y^{2}}{1-y^{2}}\right)+\frac{1}{2} \operatorname{cosec}^{-1}\left(\frac{1+z^{2}}{2 z}\right)=\picot−1(x1)+21sec−1(1−y21+y2)+21cosec−1(2z1+z2)=π.
costan−1sincot−1(x)=? \cos \tan ^{-1} \sin \cot ^{-1}(\mathrm{x})=? costan−1sincot−1(x)=?
f(x)=cot(π2−x) এবং g(x)=sin−1x f(x)=\cot \left(\frac{\pi}{2}-x\right) \text { এবং } g(x)=\sin ^{-1} x f(x)=cot(2π−x) এবং g(x)=sin−1x
A=sec−15,B=12sin−1π4,C=sin−1r A=\sec ^{-1} \sqrt{5}, B=\frac{1}{2} \sin ^{-1} \frac{\pi}{4}, C=\sin ^{-1} r A=sec−15,B=21sin−14π,C=sin−1r এবং g(x)=sinx g(x)=\sin x g(x)=sinx
