বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
যদিtan−12+tan−13+tan−14=θ \tan^{- 1}{2} + \tan^{- 1}{3} + \tan^{- 1}{4} = \theta tan−12+tan−13+tan−14=θহয়, tanθ=?\tanθ=?tanθ=?
9
7/2
3/5
4/5
tan−12+tan−13+tan−14=θ⇒tan−1(2+3+4−2×3×41−2×3−3×4−2×4)=θ⇒tan−1(35)=θtanθ=35 \begin{aligned} & \tan ^{-1} 2+\tan ^{-1} 3+\tan ^{-1} 4=\theta \\ \Rightarrow & \tan ^{-1}\left(\frac{2+3+4-2 \times 3 \times 4}{1-2 \times 3-3 \times 4-2 \times 4}\right)=\theta \\ \Rightarrow & \tan ^{-1}\left(\frac{3}{5}\right)=\theta \\ & \quad \tan \theta=\frac{3}{5}\end{aligned} ⇒⇒tan−12+tan−13+tan−14=θtan−1(1−2×3−3×4−2×42+3+4−2×3×4)=θtan−1(53)=θtanθ=53
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?
