বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
tan1/2 ( tan-1x + tan-1 1/x) = ?
x-1/2x
tan x-1/2x
1
2
tan12(tan−1x+tan−11x)=tan12(tan−1x+1x1−x1x)=tan12(tan−1x+1x0) \begin{aligned} & \tan \frac{1}{2}\left(\tan ^{-1} x+\tan ^{-1} \frac{1}{x}\right) \\ = & \tan \frac{1}{2}\left(\tan ^{-1} \frac{x+\frac{1}{x}}{1-x \frac{1}{x}}\right) \\ = & \tan \frac{1}{2}\left(\tan ^{-1} \frac{x+\frac{1}{x}}{0}\right)\end{aligned} ==tan21(tan−1x+tan−1x1)tan21(tan−11−xx1x+x1)tan21(tan−10x+x1)
=tan(12×π2)=tanπ4=1 \begin{array}{l}=\tan \left(\frac{1}{2} \times \frac{\pi}{2}\right) \\ =\tan \frac{\pi}{4} \\ =1\end{array} =tan(21×2π)=tan4π=1
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.
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
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
4(sin-1 1/√5 + cot-13) = কত?
