বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
2tan−1 15+tan−1 14=?2\tan^{-1}\ \frac{1}{5}+\tan^{-1}\ \frac{1}{4}=?2tan−1 51+tan−1 41=?
tan−1 1213\tan^{-1}\ \frac{12}{13}tan−1 1312
45°45\degree45°
tan−1 3243\tan^{-1}\ \frac{32}{43}tan−1 4332
60°60\degree60°
2tan−115+tan−114tan−1(2×151−(15)2)+tan−1(14)=tan−1(512)+tan−1(14)=tan−1(512+141−512×14)=tan−1(3243) \begin{array}{l} 2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{4} \\ \tan ^{-1}\left(\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{4}\right) \\ =\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{4}\right) \\ =\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{1}{4}}{1-\frac{5}{12} \times \frac{1}{4}}\right) \\ =\tan ^{-1}\left(\frac{32}{43}\right)\end{array} 2tan−151+tan−141tan−1(1−(51)22×51)+tan−1(41)=tan−1(125)+tan−1(41)=tan−1(1−125×41125+41)=tan−1(4332)
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.
tan−1x+tan−1y=?\tan ^{-1} x+\tan ^{-1} y = ? tan−1x+tan−1y=?
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
sin cot−1tan cos−1x\sin\ \cot^{-1}\tan\ \cos^{-1}xsin cot−1tan cos−1x এর মান কত?
