বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
f(a)=tan−1a,g(a)=sina f(a)=\tan ^{-1} a, g(a)=\sin a f(a)=tan−1a,g(a)=sina
f(13)+f(15) \mathbf{f}\left(\frac{1}{3}\right)+\mathbf{f}\left(\frac{1}{5}\right) f(31)+f(51) এর মান নির্ণয় কর।
দেখাও যে, 2f(x−yx+ytanθ2)=sec−1x+yg(π2−θ)y+xg(π2−θ) 2 f\left(\sqrt{\frac{x-y}{x+y}} \tan \frac{\theta}{2}\right)=\sec ^{-1} \frac{x+y g\left(\frac{\pi}{2}-\theta\right)}{y+x g\left(\frac{\pi}{2}-\theta\right)} 2f(x+yx−ytan2θ)=sec−1y+xg(2π−θ)x+yg(2π−θ)
সমাধান কর: g(π2−x)+g(x)=12 \mathrm{g}\left(\frac{\pi}{2}-\mathrm{x}\right)+\mathrm{g}(\mathrm{x})=\frac{1}{\sqrt{2}} g(2π−x)+g(x)=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
