বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
sin(tan-11/2+cot-13)=কত?
0
1
1/√2
3/4
sin(tan−112+tan−113)⇒sin(tan−1((1/2+1/3)/(1−1/2×1/3))⇒sin(tan−11)=12 \begin{aligned} & \sin \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}\right) \\ \Rightarrow sin(tan^{-1}((1/2+1/3)/(1-1/2×1/3)) \\ \Rightarrow & \sin \left(\tan ^{-1} 1\right) \\ = & \frac{1}{\sqrt{2}}\end{aligned} ⇒sin(tan−1((1/2+1/3)/(1−1/2×1/3))⇒=sin(tan−121+tan−131)sin(tan−11)21
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?
